1063 Set Similarity
n个序列分别先放进集合里去重。在询问的时候，遍历A集合中每个数，判断下该数在B集合中是否存在，统计存在个数（分子），分母就是两个集合大小减去分子。
1 // 1063 Set Similarity
2 #include <set>
3 #include <map>
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 set <int> se[55];
10
11 int main() {
12 int n, k, m;
13 scanf("%d", &n);
14 for(int i = 1; i <= n; i++) {
15 scanf("%d", &m);
16 while(m--) {
17 scanf("%d", &k);
18 se[i].insert(k);
19 }
20 }
21 scanf("%d", &k);
22 while(k--) {
23 int id1, id2;
24 double cnt = 0, total = 0;
25 scanf("%d %d", &id1, &id2);
26 for(int i : se[id1]) {
27 if(se[id2].find(i) != se[id2].end()) cnt++;
28 }
29 total = se[id1].size() + se[id2].size() - cnt;
30 printf("%.1f%\n", cnt * 100 / total);
31 }
32 return 0;
33 }
1064 Complete Binary Search Tree
按照中序遍历建树。对应根节点依次填入排序好的数即可。
1 // 1064 Complete Binary Search Tree
2 #include <queue>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1234;
9 int a[N], level[N], n, cnt = 0;
10
11 void solve(int root) {
12 if(root > n) return ;
13 solve(2 * root);
14 level[root] = a[++cnt];
15 solve(2 * root + 1);
16 }
17
18 int main() {
19 scanf("%d", &n);
20 for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
21 sort(a + 1, a + 1 + n);
22 solve(1);
23 for(int i = 1; i <= n; i++) {
24 if(i != 1) printf(" ");
25 printf("%d", level[i]);
26 }
27 return 0;
28 }
1065 A+B and C (64bit)
对溢出的情况特判一下。
1 // 1065 A+B and C (64bit)
2 #include <iostream>
3 using namespace std;
4
5 typedef long long ll;
6
7 int main() {
8 int t;
9 scanf("%d", &t);
10 for(int i = 1; i <= t; i++) {
11 ll a, b, c, sum = 0;
12 scanf("%lld %lld %lld", &a, &b, &c);
13 sum = a + b;
14 if(a > 0 && b > 0 && sum <= 0) {
15 printf("Case #%d: true\n", i);
16 } else if(a < 0 && b < 0 && sum >= 0) {
17 printf("Case #%d: false\n", i);
18 } else if(sum > c) {
19 printf("Case #%d: true\n", i);
20 } else {
21 printf("Case #%d: false\n", i);
22 }
23 }
24 return 0;
25 }
1066 Root of AVL Tree
AVL模板题
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 typedef struct node;
5 typedef node * tree;
6
7 struct node{
8 int v, heigh;
9 tree L,R;
10 };
11
12 int getheigh(tree root){
13 if(root==NULL) return 0;
14 return root->heigh;
15 }
16
17 void updataheigh(tree root){
18 root->heigh=max(getheigh(root->L),getheigh(root->R))+1;
19 }
20
21 int getBalance(tree root){
22 return getheigh(root->L)-getheigh(root->R);
23 }
24
25 void L(tree &root){
26 tree temp;
27 temp=root->R;
28 root->R=temp->L;
29 temp->L=root;
30 updataheigh(root);
31 updataheigh(temp);
32 root=temp;
33 }
34
35 void R(tree &root){
36 tree temp;
37 temp=root->L;
38 root->L=temp->R;
39 temp->R=root;
40 updataheigh(root);
41 updataheigh(temp);
42 root=temp;
43 }
44
45 void insertt(tree &root,int v){
46 if(root==NULL){
47 root=new node;
48 root->v=v;
49 root->heigh=1;
50 root->L=root->R=NULL;
51 return;
52 }
53 if(v<root->v){
54 insertt(root->L,v);
55 updataheigh(root);
56 if(getBalance(root)==2){
57 if(getBalance(root->L)==1){
58 R(root);
59 }
60 else if(getBalance(root->L)==-1){
61 L(root->L);
62 R(root);
63 }
64 }
65 }
66 else{
67 insertt(root->R,v);
68 updataheigh(root);
69 if(getBalance(root)==-2){
70 if(getBalance(root->R)==-1){
71 L(root);
72 }
73 else if(getBalance(root->R)==1){
74 R(root->R);
75 L(root);
76 }
77 }
78 }
79
80 }
81
82 int main(){
83 int n;
84 scanf("%d",&n);
85 int x;
86 tree root;
87 root=NULL;
88 for(int i=0;i<n;i++){
89 scanf("%d",&x);
90 insertt(root,x);
91 }
92 printf("%d\n",root->v);
93 return 0;
94 }
1067 Sort with Swap(0, i)
看到表排序的知识，摘录下。
表排序思想：N个数字的排列由若干个独立的环组成
分成以下三种环：
1. 环内只有一个元素，即该数已经在正确的位置上，不需要交换。
2. 环内有 n 个元素，含 0，需要交换次数是：n-1 （每一个非 0 元素都要和 0 换一下）
3. 环内有 n 个元素，不含 0，需要交换的次数是：n+1（把 0 换入换出 + 其余 n-1 个非 0 元素和 0 换一下）
1 // 1067 Sort with Swap(0, i)
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 const int N = 1e5 + 10;
8 bool vis[N];
9 int pos[N];
10
11 int main() {
12 int n, u, ans = 0;
13 scanf("%d", &n);
14 for(int i = 0; i < n; i++) {
15 scanf("%d", &u);
16 pos[u] = i;
17 }
18 for(int i = 0; i < n; i++) {
19 if(pos[i] != i && !vis[i]) {
20 int cnt = 0, j = i;
21 do {
22 cnt++;
23 vis[j] = true;
24 if(j == 0) cnt -= 2;
25 j = pos[j];
26 }while(j != i);
27 cnt++;
28 ans += cnt;
29 }
30 }
31 printf("%d\n", ans);
32 return 0;
33 }
1068 Find More Coins
因题目特殊的输出要求，用01背包的时候需要先用大的数填，不然如果先用小的数填，小的数会影响后面的数，最后从小的往大的拿出去即可。
1 // 1068 Find More Coins
2 #include <vector>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e4 + 10;
9 int n, m;
10 int a[N], dp[N];
11 vector <int> ans;
12 bool check[N][123];
13
14 int main() {
15 scanf("%d %d", &n, &m);
16 for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
17 sort(a + 1, a + 1 + n);
18 for(int i = n; i >= 1; i--) {
19 for(int j = m; j >= a[i]; j--) {
20 if(dp[j - a[i]] + a[i] >= dp[j]) {
21 dp[j] = dp[j - a[i]] + a[i];
22 check[i][j] = 1;
23 }
24 }
25 }
26 if(dp[m] != m) {
27 printf("No Solution\n");
28 return 0;
29 }
30 int id = 1;
31 while(m > 0) {
32 if(check[id][m]) {
33 ans.push_back(a[id]);
34 m -= a[id];
35 }
36 id++;
37 }
38 for(int i = 0; i < ans.size(); i++) {
39 if(i != 0) printf(" ", ans[i]);
40 printf("%d", ans[i]);
41 }
42 return 0;
43 }
1069 The Black Hole of Numbers
按题目模拟即可。
1 // 1069 The Black Hole of Numbers
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 int a[5];
9 bool cmp(int a, int b) {return a > b;}
10
11 int cal1(int n) {
12 memset(a, 0, sizeof(a));
13 int cnt = 0, res = 0;
14 while(n) {
15 a[++cnt] = n % 10;
16 n /= 10;
17 }
18 sort(a + 1, a + 1 + 4);
19 for(int i = 4; i >= 1; i--) {
20 res = res * 10 + a[i];
21 }
22 return res;
23 }
24
25 int cal2() {
26 int res = 0;
27 sort(a + 1, a + 1 + 4);
28 for(int i = 1; i <= 4; i++) {
29 res = res * 10 + a[i];
30 }
31 return res;
32
33 }
34
35 int main() {
36 int n;
37 scanf("%d", &n);
38 while(1) {
39 int p1 = cal1(n);
40 int p2 = cal2();
41 n = p1 - p2;
42 printf("%04d - %04d = %04d\n", p1, p2, n);
43 if(n == 6174 || n == 0) return 0;
44 }
45 return 0;
46 }
1070 Mooncake
按每吨多少钱从大到小排序下，再取即可。
1 // 1070 Mooncake
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 const int N = 1234;
8
9 struct node{
10 double storage, price;
11 }moon[N];
12
13 bool cmp(node x, node y) {
14 return x.price * y.storage > y.price * x.storage;
15 }
16
17 int main() {
18 int n;
19 double d, ans = 0;
20 scanf("%d %lf", &n, &d);
21 for(int i = 1; i <= n; i++) scanf("%lf", &moon[i].storage);
22 for(int i = 1; i <= n; i++) scanf("%lf", &moon[i].price);
23 sort(moon + 1, moon + 1 + n, cmp);
24 for(int i = 1; i <= n; i++) {
25 if(d >= moon[i].storage) {
26 ans += moon[i].price;
27 d -= moon[i].storage;
28 } else {
29 ans += moon[i].price * (1.0 * d / moon[i].storage);
30 break;
31 }
32 }
33 printf("%.2f\n", ans);
34 return 0;
35 }
1071 Speech Patterns
用map存下。更新答案即可。
1 // 1071 Speech Patterns
2 #include <map>
3 #include <string>
4 #include <iostream>
5 using namespace std;
6
7 string in, tmp, ans1;
8 map<string, int> mp;
9 map<string, int>::iterator it;
10
11 int main(){
12 int ans2 = 0;
13 getline(cin, in);
14 for(int i = 0 ; i < in.size(); i++) {
15 if(isalnum(in[i])) {
16 tmp += tolower(in[i]);
17 if(i == in.size() - 1) mp[tmp]++;
18 } else {
19 if(tmp.size()) mp[tmp]++;
20 tmp.clear();
21 }
22 }
23 for(it = mp.begin(); it!=mp.end(); it++){
24 if(ans2 < it->second){
25 ans2 = it -> second;
26 ans1 = it -> first;
27 }else if(ans2 == it -> second && ans1 > it -> first){
28 ans1 = it -> first;
29 }
30 }
31 cout << ans1 << " " << ans2 << endl;
32 }
1072 Gas Station
以加油站为起点，跑最短路即可。
1 //1072 Gas Station
2 #include <queue>
3 #include <cstdio>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N=1e3+100;
10 const int INF=0x3f3f3f3f;;
11 int n,m,k,ds,mp[N][N],dis[N],vis[N];
12
13 int toint(char s[]){
14 int sum=0;
15 int len=strlen(s);
16 for(int i=0;i<len;i++){
17 if(s[i]=='G') continue;
18 sum=sum*10+s[i]-'0';
19 }
20 if(s[0]=='G') return sum+n;
21 return sum;
22 }
23
24 void Dijkstra(int v){
25 fill(vis,vis+N,false);
26 fill(dis,dis+N,INF);
27 for(int i=1;i<=n+m;i++) dis[i]=mp[v][i];
28 dis[v]=0;
29 for(int i=0;i<n+m;i++){
30 int u=-1,minn=INF;
31 for(int j=1;j<=n+m;j++){
32 if(!vis[j]&&minn>dis[j]){
33 u=j;minn=dis[j];
34 }
35 }
36 vis[u]=true;
37 for(int j=1;j<=n+m;j++){
38 if(!vis[j]&&dis[j]>mp[u][j]+dis[u]){
39 dis[j]=mp[u][j]+dis[u];
40 }
41 }
42 }
43 }
44 int main(){
45 int num,a,b;
46 char s1[5],s2[5];
47 scanf("%d %d %d %d",&n,&m,&k,&ds);
48 fill(mp[0],mp[0]+N*N,INF);
49 for(int i=0;i<k;i++){
50 scanf("%s %s %d",s1,s2,&num);
51 a=toint(s1),b=toint(s2);
52 mp[a][b]=mp[b][a]=num;
53 }
54 int id=-1;
55 double maxDis=-1,minAvg=INF;
56 for(int i=n+1;i<=n+m;i++){
57 Dijkstra(i);
58 double avg=0;
59 double Min=INF;
60 bool flag=false;
61 for(int j=1;j<=n;j++){
62 if(dis[j]>ds){
63 flag=true;
64 break;
65 }
66 if(Min>dis[j]){
67 Min=dis[j];
68 }
69 avg+=1.0*dis[j]/n;
70 }
71 if(flag) continue;
72 if(maxDis<Min){
73 maxDis=Min;
74 id=i;
75 minAvg=avg;
76 }
77 else if(maxDis==Min&&minAvg>avg){
78 id=i;
79 minAvg=avg;
80 }
81
82 }
83 if(id==-1) printf("No Solution\n");
84 else{
85 printf("G%d\n",id-n);
86 printf("%.1f %.1f\n",maxDis,minAvg);
87 }
88 return 0;
89 }
1073 Scientific Notation
按题意模拟即可。
1 // 1073 Scientific Notation
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e4 + 10;
9 char s[N];
10
11 int main() {
12 int n, i = 0;
13 scanf("%s", s);
14 n = strlen(s);
15 while(s[i] != 'E') {
16 i++;
17 }
18 i++;
19 if(s[0] != '+') printf("%c", s[0]);
20 int m = 0;
21 for(int j = i + 1; j < n; j++) m = 10 * m + (s[j] - '0');
22 if(s[i] == '+') {
23 m+=2;
24 for(int j = 1; j < i - 1; j++) {
25 m--;
26 if(m == -1) printf(".");
27 if(s[j] != '.') printf("%c", s[j]);
28 }
29 while(m > 0) {
30 printf("0");
31 m--;
32 }
33 } else {
34 m--;
35 printf("0.");
36 while(m > 0) {
37 printf("0");
38 m--;
39 }
40 for(int j = 1; j < i - 1; j++) {
41 if(s[j] != '.') printf("%c", s[j]);
42 }
43 }
44 return 0;
45 }
1074 Reversing Linked List
每k个长度翻转一次。$j$位置对应交换的位置为$(2 * i + k - j - 1)$，这个只使用前半和后半交换，注意特判跳出。
1 //1074 Reversing Linked List
2 #include <cstdio>
3 #include <vector>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e5 + 10;
9 int nxt[N], val[N], res[N];
10
11 int main() {
12 int root, n, k, cnt = 0;
13 scanf("%d%d%d", &root, &n, &k);
14 for(int i = 1; i <= n; i++) {
15 int u, v, w;
16 scanf("%d%d%d", &u, &v, &w);
17 val[u] = v; nxt[u] = w;
18 }
19 while(root != -1) {
20 res[cnt++] = root;
21 root = nxt[root];
22 }
23 for(int i = 0; i + k <= cnt; i += k) {
24 for(int j = i; j < (i + k); j++) {
25 if(j >= (2 * i + k - j - 1)) break;
26 swap(res[(2 * i + k - j - 1)], res[j]);
27 }
28 }
29 for(int i = 0; i < cnt - 1; i++) {
30 printf("%05d %d %05d\n", res[i], val[res[i]], res[i + 1]);
31 }
32 printf("%05d %d -1\n", res[cnt - 1], val[res[cnt - 1]]);
33 return 0;
34 }
1075 PAT Judge
结构体排序，注意一下区别提交得0分和提交没有通过编译。
1 // 1075 PAT Judge
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e5 + 10;
9 int v[6];
10 int n, k, m;
11 struct node{
12 int q[6], id, sum, ok, perfect;
13 }p[N];
14
15 void init() {
16 for(int i = 1; i <= n; i++)
17 for(int j = 1; j <= k; j++)
18 p[i].q[j] = -2, p[i].id = i;
19 }
20
21 bool cmp(node x, node y) {
22 if(x.sum == y.sum) {
23 if(x.perfect == y.perfect) return x.id < y.id;
24 return x.perfect > y.perfect;
25 }
26 return x.sum > y.sum;
27 }
28
29 int main() {
30 int r = 1;
31 scanf("%d %d %d", &n, &k, &m);
32 for(int i = 1; i <= k; i++) scanf("%d", &v[i]);
33 init();
34 for(int i = 1; i <= m; i++) {
35 int uid, pid, val;
36 scanf("%d %d", &uid, &pid);
37 scanf("%d", &val);
38 p[uid].q[pid] = max(p[uid].q[pid], val);
39 }
40 for(int i = 1; i <= n; i++) {
41 int cnt = 0, perfect = 0;
42 bool ok = 0;
43 for(int j = 1; j <= k; j++) {
44 if(p[i].q[j] != -1 && p[i].q[j] != -2) cnt += p[i].q[j], ok = 1;
45 if(p[i].q[j] == v[j]) perfect++;
46 }
47 if(ok) p[i].ok = 1;
48 else p[i].ok = 0;
49 p[i].sum = cnt;
50 p[i].perfect = perfect;
51 }
52 sort(p + 1, p + 1 + n, cmp);
53 for(int i = 1; i <= n; i++) {
54 if(p[i].ok == 0) continue;
55 if(p[i].sum != p[i - 1].sum) r = i;
56 printf("%d %05d %d", r, p[i].id, p[i].sum);
57 for(int j = 1; j <= k; j++) {
58 if(p[i].q[j] == -2) printf(" -");
59 else if(p[i].q[j] == -1) printf(" 0");
60 else printf(" %d",p[i].q[j]);
61 }
62 printf("\n");
63 }
64 return 0;
65 }
1076 Forwards on Weibo
DFS或BFS，DFS的时候注意剪枝。
DFS版本
1 //1076 Forwards on Weibo
2 #include <vector>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1234;
9 vector <int> E[N];
10 bool vis[N];
11 int layer[N];
12 int n, m, k, u, sz, ans;
13
14 void dfs(int u, int l) {
15 vis[u] = 1;
16 layer[u] = l;
17 if(l <= k) {
18 for(int i = 0; i < E[u].size(); i++) {
19 int v = E[u][i];
20 if(!vis[v] || layer[v] > l + 1) {
21 dfs(v, l + 1);
22 }
23 }
24 }
25 }
26
27 int main() {
28 scanf("%d %d", &n, &k);
29 for(int i = 1; i <= n; i++) {
30 scanf("%d", &sz);
31 while(sz) {
32 sz--;
33 scanf("%d", &u);
34 E[u].push_back(i);
35 }
36 }
37 scanf("%d", &m);
38 while(m--) {
39 for(int i = 1; i <= n; i++) vis[i] = 0, layer[i] = 0;
40 scanf("%d", &u);
41 ans = 0;
42 dfs(u, 1);
43 for(int i = 1; i <= n; i++) if(layer[i] > 0) ans++;
44 printf("%d\n", ans - 1);
45 }
46 return 0;
47 }BFS版本
1 //1076 Forwards on Weibo
2 #include <queue>
3 #include <vector>
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 1234;
10 vector <int> E[N];
11 bool vis[N];
12 int n, m, k, u, sz;
13
14 void bfs() {
15 int ans = 0;
16 for(int i = 1; i <= n; i++) vis[i] = 0;
17 queue < pair<int,int> > Q;
18 Q.push(make_pair(u, 0));
19 vis[u] = 1;
20 while(!Q.empty()) {
21 int cur = Q.front().first;
22 int layer = Q.front().second;
23 Q.pop();
24 if(layer >= k) continue;
25 for(int i = 0; i < E[cur].size(); i++) {
26 int v = E[cur][i];
27 if(!vis[v]) {
28 ans++;
29 vis[v] = 1;
30 Q.push(make_pair(v, layer + 1));
31 }
32 }
33 }
34 printf("%d\n", ans);
35 }
36
37 int main() {
38 scanf("%d %d", &n, &k);
39 for(int i = 1; i <= n; i++) {
40 scanf("%d", &sz);
41 while(sz) {
42 sz--;
43 scanf("%d", &u);
44 E[u].push_back(i);
45 }
46 }
47 scanf("%d", &m);
48 while(m--) {
49 scanf("%d", &u);
50 bfs();
51 }
52 return 0;
53 }
1077 Kuchiguse
模拟
1 //1077 Kuchiguse
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 int id[110];
8 char s[110][260];
9
10 int main() {
11 int n, cnt = 1;
12 scanf("%d", &n);
13 getchar();
14 for(int i = 1; i <= n; i++) {
15 char c;
16 while(scanf("%c", &c) != EOF && c != '\n') {
17 s[i][id[i]++] = c;
18 }
19 }
20 while(1) {
21 for(int i = 2; i <= n; i++) {
22 if( ((id[i] - cnt) < 0) || (s[i][id[i] - cnt] != s[1][id[1] - cnt]) ) {
23 cnt--;
24 if(cnt == 0) printf("nai");
25 else {
26 for(int i = id[1] - cnt; i < id[1] ; i++) {
27 printf("%c", s[1][i]);
28 }
29 }
30 return 0;
31 }
32 }
33 cnt++;
34 }
35 return 0;
36 }
1078 Hashing
Hash二次探测。0^2，1^2，2^2...。
1 //1078 Hashing
2 #include <iostream>
3 using namespace std;
4
5 const int N = 1e4 + 10;
6 bool vis[N];
7
8 bool check(int x) {
9 if(x < 2) return false;
10 for(int i = 2; i * i <= x; i++) {
11 if(x % i == 0) return false;
12 }
13 return true;
14 }
15
16 int main() {
17 int n, m, p;
18 cin >> m >> n;
19 while(!check(m)) {m++;}
20 for(int i = 1; i <= n; i++) {
21 bool f = 0;
22 cin >> p;
23 if(i != 1) printf(" ");
24 for(int i = 0; i < m; i++) {
25 int tmp = (p + i * i) % m;
26 if(!vis[tmp]) {
27 f = 1;
28 printf("%d", tmp);
29 vis[tmp] = 1;
30 break;
31 }
32 }
33 if(!f) printf("-");
34 }
35 return 0;
36 }
1079 Total Sales of Supply Chain
从根跑一遍DFS到叶子节点。统计每个叶子节点的能够产生的销售额。
1 //1079 Total Sales of Supply Chain
2 #include <vector>
3 #include <cstdio>
4 using namespace std;
5
6 const int N = 1e5 + 10;
7 int cnt[N];
8 bool vis[N];
9 double p, r;
10 double price[N];
11 vector <int> E[N];
12
13 void dfs(int u, double val) {
14 vis[u] = 1; price[u] = val;
15 for(int i = 0; i < E[u].size(); i++) {
16 int v = E[u][i];
17 if(!vis[v]) dfs(v, val * (1 + r / 100));
18 }
19 }
20
21 int main() {
22 int n, k, u;
23 scanf("%d %lf %lf", &n, &p, &r);
24 for(int i = 0; i < n; i++) {
25 scanf("%d", &k);
26 if(k == 0){
27 scanf("%d", &cnt[i]);
28 } else {
29 while(k--) {
30 scanf("%d", &u);
31 E[i].push_back(u);
32 }
33 }
34 }
35 double ans = 0;
36 dfs(0, p);
37 for(int i = 0; i < n; i++) {
38 ans += price[i] * cnt[i];
39 }
40 printf("%.1f\n", ans);
41 return 0;
42 }
1080 Graduate Admission
按照题意模拟，用set从小到大排序特性会方便很多。
1 // 1080 Graduate Admission
2 #include <set>
3 #include <cstdio>
4 #include <vector>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 4e4 + 10;
10 struct Stu{
11 int Id, Ge, Gi, Gh, Rank;
12 int choices[10];
13 };
14
15 struct Sch{
16 int maxn, lastRank;
17 set<int> admit;
18 };
19
20 bool cmp(Stu x, Stu y){
21 if(x.Gh == y.Gh) return x.Ge > y.Ge;
22 return x.Gh > y.Gh;
23 }
24
25 set<int>::iterator it;
26
27 int main(){
28 int n, m, k;
29 scanf("%d%d%d", &n, &m, &k);
30 vector<Sch> sch(m);
31 vector<Stu> stu(n);
32 for(int i = 0; i < m; i++) scanf("%d", &sch[i].maxn);
33 for(int i = 0; i < n; i++){
34 scanf("%d%d", &stu[i].Ge, &stu[i].Gi);
35 stu[i].Id=i; stu[i].Gh = stu[i].Ge + stu[i].Gi;
36 for(int j = 0; j < k; j++) scanf("%d", &stu[i].choices[j]);
37 }
38 sort(stu.begin(), stu.end(), cmp);
39 int Rank = 1;
40 for(int i = 0; i < n; i++){
41 if(i==0 || (stu[i].Gh == stu[i-1].Gh && stu[i].Gi == stu[i-1].Gi)) stu[i].Rank = Rank;
42 else{
43 Rank = i + 1;
44 stu[i].Rank = Rank;
45 }
46 }
47 for(int i = 0; i < n; i++){
48 bool f=0;
49 for(int j = 0; j < k; j++){
50 int schid = stu[i].choices[j];
51 if(sch[schid].maxn > 0){
52 sch[schid].admit.insert(stu[i].Id);
53 sch[schid].maxn--;
54 sch[schid].lastRank = stu[i].Rank;
55 f = true;
56 }else if(sch[schid].lastRank == stu[i].Rank) {
57 sch[schid].admit.insert(stu[i].Id);
58 f = true;
59 }
60 if(f) break;
61 }
62 }
63 for(int i = 0; i < m; i++){
64 for(it = sch[i].admit.begin(); it!=sch[i].admit.end(); it++){
65 if(it == sch[i].admit.begin()) printf("%d", *it);
66 else printf(" %d", *it);
67 }
68 printf("\n");
69 }
70 return 0;
71 }
1081 Rational Sum
通分，注意最后若答案为0的时的格式输出。
1 // 1081 Rational Sum
2 #include <cstdio>
3 #include <algorithm>
4 using namespace std;
5
6 typedef long long ll;
7 ll a[123], b[123];
8
9 int main() {
10 ll n, nr = 0, dm = 1;
11 scanf("%lld", &n);
12 for(ll i = 1; i <= n; i++) {
13 scanf("%lld/%lld", &a[i], &b[i]);
14 dm = b[i] / __gcd(b[i], dm) * dm;
15 }
16 for(ll i = 1; i<= n; i++) {
17 nr = nr + (dm / b[i]) * a[i];
18 }
19 ll g = __gcd(nr, dm);
20 if(g != 0) nr /= g, dm /= g;
21 if(nr / dm) {
22 printf("%lld", nr / dm);
23 nr = nr - (nr / dm) * dm;
24 if(nr != 0) printf(" %lld/%lld", nr, dm);
25 return 0;
26 }
27 nr = nr - (nr / dm) * dm;
28 if(nr != 0) printf("%lld/%lld", nr, dm);
29 else printf("0");
30 return 0;
31 }
1082 Read Number in Chinese
模拟，注意细节即可。
1 // 1082 Read Number in Chinese
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7
8 string a[10] = {"ling","yi","er","san","si","wu","liu","qi","ba","jiu"};
9 string b[10] = {"","Shi","Bai","Qian","Wan","Shi","Bai","Qian","Yi"};
10
11 int main() {
12 int n, f = 0;
13 string s;
14 cin >> s;
15 if(s[0] == '-') {
16 cout << "Fu" << " ";
17 s = s.substr(1);
18 }
19 n = s.size();
20 reverse(s.begin(), s.end());
21 for(int i = n - 1; i >= 1; i--) {
22 if(i == n - 1) {
23 cout << a[s[i] - '0'] << " " << b[i];
24 } else {
25 if(s[i] != '0') {
26 if(f) {
27 cout << " " << a[0];
28 f = 0;
29 }
30 cout << " " << a[s[i] - '0'] << " " << b[i];
31 } else {
32 f = 1;
33 if(i == 4) cout << " " << b[4];
34 }
35 }
36 }
37 if(f && s[0] != '0') {
38 cout << " " << a[0];
39 }
40 if(s.size() == 1) {
41 cout << a[s[0] - '0'];
42 }
43 else if(s.size() > 1 && s[0] != '0'){
44 cout << " " << a[s[0] - '0'];
45 }
46 return 0;
47 }
1083 List Grades
结构体排序
1 // 1083 List Grades
2 #include <cstdio>
3 #include <algorithm>
4 using namespace std;
5
6 const int N = 2e5 + 10;
7 struct node {
8 char name[15], id[15];
9 int grade;
10 }recoder[N];
11
12 bool cmp(node x, node y) {
13 return x.grade > y.grade;
14 }
15
16 int main() {
17 bool f = 0;
18 int n;
19 scanf("%d", &n);
20 for(int i = 1; i <= n; i++) {
21 scanf("%s %s %d", recoder[i].name, recoder[i].id, &recoder[i].grade);
22 }
23 int g1, g2;
24 scanf("%d%d", &g1, &g2);
25 sort(recoder + 1, recoder + 1 + n, cmp);
26 for(int i = 1; i <= n; i++) {
27 if(recoder[i].grade <= g2 && recoder[i].grade >= g1) {
28 printf("%s %s\n", recoder[i].name, recoder[i].id);
29 f = 1;
30 }
31 }
32 if(!f) printf("NONE\n");
33 return 0;
34 }map标记一下。
1 // 1084 Broken Keyboard
2 #include <map>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 map <char, int> m;
9
10 int main() {
11 char c;
12 string s1, s2;
13 cin >> s1 >> s2;
14 for(int i = 0; i < s2.size(); i++) {
15 m[s2[i]] = 1;
16 if(s2[i] >= 'a' && s2[i] <= 'z') {
17 c = s2[i] - 32;
18 m[c] = 1;
19 }
20 else if(s2[i] >= 'A' && s2[i] <= 'Z') {
21 c = s2[i] + 32;
22 m[c] = 1;
23 }
24 }
25 for(int i = 0; i < s1.size(); i++) {
26 if(!m[s1[i]]) {
27 m[s1[i]] = 1;
28 if(s1[i] >= 'a' && s1[i] <= 'z') {
29 c = s1[i] - 32;
30 m[c] = 1;
31 cout << c;
32 }
33 else if(s1[i] >= 'A' && s1[i] <= 'Z') {
34 c = s1[i] + 32;
35 m[c] = 1;
36 cout << s1[i];
37 } else {
38 cout << s1[i];
39 }
40 }
41 }
42 return 0;
43 }
1085 Perfect Sequence
二分搜索
1 // 1085 Perfect Sequence
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 typedef long long ll;
8 const int N = 1e5 + 10;
9 ll a[N];
10
11 int main() {
12 int ans = 0;
13 ll n, p;
14 scanf("%lld %lld", &n, &p);
15 for(int i = 0; i < n; i++) scanf("%lld", &a[i]);
16 sort(a, a + n);
17 for(int i = 0; i < n; i++) {
18 int pos1 = lower_bound(a, a + n, a[i]) - a;
19 int pos2 = lower_bound(a, a + n, a[i] * p + 1) - a;
20 pos2--;
21 ans = max(ans, pos2 - pos1 + 1);
22 }
23 printf("%d\n", ans);
24 return 0;
25 }
1086 Tree Traversals Again
由题目可以得知push的时候为往下找，分情况讨论存左孩子还是右孩子，pop的时候表示往右走，下一次push存右孩子，最后后序遍历一下即可。
1 // 1086 Tree Traversals Again
2 #include <stack>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 char s[5];
9 struct node{
10 int lch, rch;
11 }tree[35];
12 stack <int> sta;
13 int n, u, fa, root = -1, child = 0;
14
15 void postorder(int u) {
16 if(tree[u].lch != 0) postorder(tree[u].lch);
17 if(tree[u].rch != 0) postorder(tree[u].rch);
18 if(u == root) printf("%d", u);
19 else printf("%d ", u);
20 }
21
22 int main() {
23 scanf("%d", &n);
24 n *= 2;
25 while(n--) {
26 scanf("%s", s);
27 if(s[1] == 'u') {
28 scanf("%d", &u);
29 sta.push(u);
30 if(root == -1) {
31 root = u;
32 } else {
33 if(child == 0) tree[fa].lch = u;
34 else tree[fa].rch = u;
35 }
36 fa = u;
37 child = 0;
38 } else {
39 if(sta.size())
40 fa = sta.top();
41 sta.pop();
42 child = 1;
43 }
44 }
45 postorder(root);
46 return 0;
47 }
1087 All Roads Lead to Rome
先Floyd得到最短路，再DFS搜，搜的过程中更新答案。
1 // 1087 All Roads Lead to Rome
2 #include <bits/stdc++.h>
3 using namespace std;
4
5 const int N = 205;
6 int n, m, cost, cnt = 0, maxhappy = 0;
7 int happy[N], d[N][N];
8 string s[N], s1, s2;
9 map<string, int> mp;
10 vector<int> E[N], res, ans;
11 bool vis[N];
12
13 void dfs(int u, int en, int co) {
14 if(u == en && co == cost) {
15 int sum = 0;
16 for(int i = 0; i < res.size(); i++) {
17 sum += happy[res[i]];
18 }
19 if(sum > maxhappy) {
20 maxhappy = sum;
21 ans.clear();
22 for(int i = 0; i < res.size(); i++) ans.push_back(res[i]);
23 }
24 else if(sum == maxhappy && res.size() < ans.size()) {
25 ans.clear();
26 for(int i = 0; i < res.size(); i++) ans.push_back(res[i]);
27 }
28 cnt++;
29 return ;
30 }
31 for(int k = 0; k < E[u].size(); k++) {
32 int v = E[u][k];
33 if(vis[v]) continue;
34 res.push_back(v);
35 vis[v] = 1;
36 dfs(v, en, co + d[u][v]);
37 vis[v] = 0;
38 res.pop_back();
39 }
40 }
41
42 int main() {
43 cin >> n >> m >> s[1];
44 mp[s[1]] = 1;
45 memset(d, 0x3f, sizeof(d));
46 for(int i = 0; i <= n; i++) d[i][i] = 0;
47 for(int i = 2; i <= n; i++) {
48 cin >> s[i] >> happy[i];
49 mp[s[i]] = i;
50 }
51 for(int i = 1; i <= m; i++) {
52 cin >> s1 >> s2 >> cost;
53 d[mp[s1]][mp[s2]] = cost;
54 d[mp[s2]][mp[s1]] = cost;
55 E[mp[s2]].push_back(mp[s1]);
56 E[mp[s1]].push_back(mp[s2]);
57 }
58
59 for(int k = 1; k <= n; k++)
60 for(int i = 1; i <= n; i++)
61 for(int j = 1; j <= n; j++)
62 d[i][j] = min(d[i][j] ,d[i][k] + d[k][j]);
63
64 cost = d[mp[s[1]]][mp["ROM"]];
65 vis[mp[s[1]]] = 1;
66 dfs(mp[s[1]], mp["ROM"], 0);
67
68 cout << cnt << " " << cost << " " << maxhappy << " " << maxhappy / ans.size() << endl;
69 cout << s[1];
70 for(int i = 0; i < ans.size(); i++) cout << "->" << s[ans[i]];
71 return 0;
72 }
1088 Rational Arithmetic
注意相乘会爆long long，判断负可以分别判断x 和 y
1 // 1088 Rational Arithmetic
2 #include <bits/stdc++.h>
3 using namespace std;
4
5 typedef long long ll;
6 ll gcd(ll a, ll b) {
7 return b == 0 ? a : gcd(b, a % b);
8 }
9
10 void func(ll x, ll y) {
11 bool f = 0;
12 if(x * y == 0) {
13 if(y == 0) printf("Inf");
14 else if(x == 0) printf("0");
15 return ;
16 }
17 if((x > 0 && y < 0) || (x < 0 && y > 0)) f = 1;
18 x = abs(x); y = abs(y);
19 ll val = x / y;
20 ll res = x - val * y;
21 if(f) printf("(-");
22 if(val != 0) printf("%lld", val);
23 if(res == 0) {
24 if(f) printf(")");
25 return ;
26 }
27 if(val != 0) printf(" ");
28 ll g = gcd(res, y);
29 res /= g; y /= g;
30 printf("%lld/%lld", res, y);
31 if(f) printf(")");
32 }
33
34 int main() {
35 ll a, b, c, d;
36 scanf("%lld/%lld %lld/%lld", &a, &b, &c, &d);
37 func(a, b);printf(" + ");func(c, d);printf(" = ");func(a * d + b * c, b * d);printf("\n");
38 func(a, b);printf(" - ");func(c, d);printf(" = ");func(a * d - b * c, b * d);printf("\n");
39 func(a, b);printf(" * ");func(c, d);printf(" = ");func(a * c, b * d);printf("\n");
40 func(a, b);printf(" / ");func(c, d);printf(" = ");func(a * d, b * c);
41 return 0;
42 }
1089 Insert or Merge
先判断是否为插入排序，不是的话，跑一遍归并，直到a数组和b数组一样，再进行一次归并操作。
1 // 1089 Insert or Merge
2 #include <iostream>
3 #include <algorithm>
4 using namespace std;
5
6 const int N = 105;
7 int a[N], b[N];
8
9 int main() {
10 int n, i, j;
11 cin >> n;
12 for (i = 0; i < n; i++) cin >> a[i];
13 for (i = 0; i < n; i++) cin >> b[i];
14 for (i = 0; i < n - 1 && b[i] <= b[i + 1]; i++);
15 for (j = i + 1; a[j] == b[j] && j < n; j++);
16 if (j == n) {
17 cout << "Insertion Sort" << endl;
18 sort(a, a + i + 2);
19 } else {
20 cout << "Merge Sort" << endl;
21 int k = 1, f = 1;
22 while(f) {
23 f = 0;
24 for (i = 0; i < n; i++) {
25 if (a[i] != b[i]) f = 1;
26 }
27 k = k * 2;
28 for (i = 0; i < n / k; i++)
29 sort(a + i * k, a + (i + 1) * k);
30 sort(a + n / k * k, a + n);
31 }
32 }
33 for(j = 0; j < n; j++) {
34 if (j != 0) cout << " ";
35 cout << a[j];
36 }
37 return 0;
38 }
1090 Highest Price in Supply Chain
DFS跑一遍树即可。
1 // 1090 Highest Price in Supply Chain
2 #include <cstdio>
3 #include <vector>
4 using namespace std;
5
6 const int N = 1e5 + 10;
7 int n, u, cnt, root;
8 double p, r, maxp = 0, sellp[N];
9 vector <int> E[N];
10
11 void dfs(int a, double price) {
12 sellp[a] = price;
13 maxp = max(maxp, price);
14 for(int i = 0; i < E[a].size(); i++) {
15 int b = E[a][i];
16 dfs(b, price * (1.0 + r / 100));
17 }
18 }
19
20 int main() {
21 scanf("%d %lf %lf", &n, &p, &r);
22 for(int i = 0; i < n; i++) {
23 scanf("%d", &u);
24 if(u == -1) root = i;
25 else E[u].push_back(i);
26 }
27 dfs(root, p);
28 for(int i = 0; i < n; i++) {
29 if(sellp[i] == maxp) cnt++;
30 }
31 printf("%.2f %d\n", maxp, cnt);
32 return 0;
33 }
1091 Acute Stroke
3维BFS裸题。
1 // 1091 Acute Stroke
2 #include <queue>
3 #include <cstdio>
4 using namespace std;
5
6 int n, m, l, t, ans = 0;
7 int mp[70][1300][130];
8 bool vis[70][1300][130];
9 struct node{
10 int z, x, y;
11 };
12 int dx[6] = {0, 0, 0, 0, -1, 1};
13 int dy[6] = {0, 0, -1, 1, 0, 0};
14 int dz[6] = {-1, 1, 0, 0, 0, 0};
15
16 bool check(int z, int x, int y) {
17 if(z >= 1 && z <= l && x >= 1 && x <= n && y >= 1 && y <= m) return true;
18 return false;
19 }
20
21 int bfs(int z, int x, int y) {
22 queue <node> q;
23 node tmp;
24 int res = 0;
25 q.push(node{z, x, y});
26 vis[z][x][y] = 1;
27 while(!q.empty()) {
28 res++;
29 tmp = q.front();
30 q.pop();
31 for(int i = 0; i < 6; i++) {
32 int nz = tmp.z + dz[i];
33 int nx = tmp.x + dx[i];
34 int ny = tmp.y + dy[i];
35 if(check(nz, nx, ny) && !vis[nz][nx][ny] && mp[nz][nx][ny]) {
36 vis[nz][nx][ny] = 1;
37 q.push(node{nz, nx, ny});
38 }
39 }
40
41 }
42 if(res >= t) return res;
43 else return 0;
44 }
45
46 int main() {
47 scanf("%d%d%d%d", &n, &m, &l, &t);
48 for(int z = 1; z <= l; z++)
49 for(int x = 1; x <= n; x++)
50 for(int y = 1; y <= m; y++)
51 scanf("%d", &mp[z][x][y]);
52
53 for(int z = 1; z <= l; z++)
54 for(int x = 1; x <= n; x++)
55 for(int y = 1; y <= m; y++)
56 if(mp[z][x][y] && !vis[z][x][y]) ans += bfs(z, x, y);
57
58 printf("%d\n", ans);
59 return 0;
60 }
1092 To Buy or Not to Buy
用数组标记一下即可。
1 // 1092 To Buy or Not to Buy
2 #include <iostream>
3 using namespace std;
4
5 int a[123];
6 string s1, s2;
7
8 int main(){
9 cin >> s1 >> s2;
10 int c1 = 0, c2 = 0;
11 for(int i = 0; i < s1.size(); i++) a[s1[i]]++;
12 for(int i = 0; i < s2.size(); i++) a[s2[i]]--;
13 for(int i = 48; i < 123; i++) {
14 if(a[i] < 0) c1+=a[i];
15 else c2+=a[i];
16 }
17 if(c1 < 0) cout << "No " << -c1 << endl;
18 else cout << "Yes " << c2 << endl;
19 return 0;
20 }
1093 Count PAT's
统计一下P,A,T。在A位置计算前面有几个P，后面有几个T。
1 // 1093 Count PAT's
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5
6 typedef long long ll;
7 const int N = 1e5 + 10;
8 const ll mod = 1000000007;
9 char s[N];
10 ll sum[N][5];
11
12 int main() {
13 int n;
14 ll ans = 0;
15 scanf("%s", s + 1);
16 n = strlen(s + 1);
17 for(int i = 1; i <= n; i++) {
18 for(int j = 1; j <= 3; j++) sum[i][j] = sum[i - 1][j];
19 if(s[i] == 'P') sum[i][1]++;
20 else if(s[i] == 'A') sum[i][2]++;
21 else sum[i][3]++;
22 }
23 for(int i = 1; i <= n; i++) {
24 if(s[i] == 'A') {
25 ans = (ans + (sum[i][1] * (sum[n][3] - sum[i][3])) % mod) % mod;
26 }
27 }
28 printf("%lld\n", ans);
29 return 0;
30 }
1094 The Largest Generation
DFS跑下树，记录下即可。
1 // 1094 The Largest Generation
2 #include <vector>
3 #include <cstdio>
4 using namespace std;
5
6 const int N = 105;
7 int n, m, cnt[N];
8 vector <int> children[N];
9
10 void dfs(int u, int level) {
11 for(int i = 0; i < children[u].size(); i++) {
12 dfs(children[u][i], level + 1);
13 }
14 cnt[level]++;
15 }
16
17 int main() {
18 scanf("%d%d", &n, &m);
19 while(m--) {
20 int r, k, c;
21 scanf("%d%d", &r, &k);
22 while(k--) {
23 scanf("%d", &c);
24 children[r].push_back(c);
25 }
26 }
27 dfs(1, 1);
28 int mx = 0, id;
29 for(int i = 1; i <= n; i++) {
30 if(cnt[i] > mx) {
31 mx = cnt[i];
32 id = i;
33 }
34 }
35 printf("%d %d\n", mx, id);
36 return 0;
37 }
1095 Cars on Campus
按同个车牌号并以时间排序，模拟过去即可。
1 // 1095 Cars on Campus
2 #include <map>
3 #include <vector>
4 #include <cstdio>
5 #include <cstring>
6 #include <iostream>
7 #include <algorithm>
8 using namespace std;
9
10 const int N = 1e4 + 5;
11 char s[N][10], op[5];
12 int t[N], f[N], id[N];
13 int cnt[3600 * 60 + 5], maxtime;
14 map<string, int> m;
15 vector <string> ans;
16
17 bool cmp(int x, int y) {
18 if(!strcmp(s[x], s[y])) {
19 return t[x] < t[y];
20 } else {
21 return strcmp(s[x], s[y]) < 0;
22 }
23 }
24
25 int main() {
26 int n, k, hh, mm, ss;
27 scanf("%d%d", &n, &k);
28 for(int i = 0; i < n; i++) {
29 id[i] = i;
30 scanf("%s %d:%d:%d %s", s[i], &hh, &mm, &ss, op);
31 t[i] = 3600 * hh + 60 * mm + ss;
32 if(op[0] == 'i') f[i] = 0;
33 else f[i] = 1;
34 }
35 sort(id, id + n, cmp);
36 for(int i = 1; i < n; i++) {
37 if(!strcmp(s[id[i]], s[id[i - 1]])) {
38 if(f[id[i - 1]] == 0 && f[id[i]] == 1) {
39 cnt[t[id[i - 1]]]++;
40 cnt[t[id[i]]]--;
41 m[s[id[i]]] = m[s[id[i]]] + t[id[i]] - t[id[i - 1]];
42 maxtime = max(maxtime, m[s[id[i]]]);
43 }
44 }
45 }
46 for(int i = 0; i <= 3600 * 60; i++) cnt[i] += cnt[i - 1];
47 while(k--) {
48 scanf("%d:%d:%d", &hh, &mm, &ss);
49 hh = 3600 * hh + 60 * mm + ss;
50 printf("%d\n", cnt[hh]);
51 }
52 for(int i = 0; i < n; i++) {
53 if(m[s[i]] == maxtime) {
54 ans.push_back(s[i]);
55 m[s[i]] = 0;
56 }
57 }
58 sort(ans.begin(), ans.end());
59 for(int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
60 hh = maxtime / 3600; maxtime -= 3600 * hh;
61 mm = maxtime / 60; maxtime -= 60 * mm;
62 ss = maxtime;
63 printf("%02d:%02d:%02d\n", hh, mm, ss);
64 return 0;
65 }
1096 Consecutive Factors
暴力找起点，然后找可以被整除的数。
1 // 1096 Consecutive Factors
2 #include <iostream>
3 #include <algorithm>
4 using namespace std;
5
6 typedef long long ll;
7
8 int main() {
9 ll n, f = 0, maxlen = 0;
10 cin >> n;
11 for(ll i = 2; i * i <= n; i++){
12 ll m = 1, j = i;
13 for(; j * j <= n; j++) {
14 m *= j;
15 if(n % m != 0) break;
16 }
17 if(j - i > maxlen) {
18 maxlen = j - i;
19 f = i;
20 }
21 }
22 if(f == 0) {
23 cout << 1 << endl << n << endl;
24 return 0;
25 }
26 cout << maxlen << endl;
27 ll tmp = 1;
28 cout << f;
29 for(ll i = f + 1; tmp < n && (i - f) < maxlen; i++) {
30 tmp *= i;
31 cout << "*" << i;
32 }
33 return 0;
34 }
1097 Deduplication on a Linked List
从根跑下树，把仍然保留的移除的都存起来。最后输出即可。
1 #include <map>
2 #include <vector>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e5 + 10;
9 int nxt[N], val[N];
10 map<int, int> m;
11
12 struct node{
13 int u, w, v;
14 };
15
16 vector <node> Remain, Remove;
17
18 int main() {
19 int root, cur, n;
20 scanf("%d %d", &root, &n);
21 for(int i = 1; i <= n; i++) {
22 int u, v, w;
23 scanf("%d %d %d", &u, &w, &v);
24 nxt[u] = v;
25 val[u] = w;
26 }
27 cur = root;
28 while(cur != -1) {
29 if(!m[abs(val[cur])]) {
30 Remain.push_back({cur, val[cur], nxt[cur]});
31 m[abs(val[cur])] = 1;
32 } else {
33 Remove.push_back({cur, val[cur], nxt[cur]});
34 }
35 cur = nxt[cur];
36 }
37 for(int i = 0; i < Remain.size(); i++) {
38 if(i != Remain.size() - 1) printf("%05d %d %05d\n", Remain[i].u, Remain[i].w, Remain[i + 1].u);
39 else printf("%05d %d -1\n", Remain[i].u, Remain[i].w);
40 }
41 for(int i = 0; i < Remove.size(); i++) {
42 if(i != Remove.size() - 1) printf("%05d %d %05d\n", Remove[i].u, Remove[i].w, Remove[i + 1].u);
43 else printf("%05d %d -1\n", Remove[i].u, Remove[i].w);
44 }
45 return 0;
46 }
1098 Insertion or Heap Sort
判断下排序类型，插入排序很容易解决，再扔个堆排序上去即可。
1 // 1098 Insertion or Heap Sort
2 #include <cstdio>
3 #include <cstring>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 105;
9 int a[N], b[N];
10
11 void solve(int l, int r) {
12 int i = 0, j = 1;
13 while(j <= r) {
14 if(j + 1 <= r && b[j] < b[j + 1]) j++;
15 if(b[j] > b[i]) {
16 swap(b[j], b[i]);
17 i = j;
18 j = 2 * (i + 1) -1;
19 } else {
20 break;
21 }
22 }
23 }
24
25 int main() {
26 int n, k, f = 1;
27 scanf("%d", &n);
28 for(int i = 0; i < n; i++) scanf("%d", &a[i]);
29 for(int i = 0; i < n; i++) scanf("%d", &b[i]);
30 for(k = 0; k < n - 1 && b[k] <= b[k + 1]; k++) ;
31 for(int i = k + 1; i < n; i++) {
32 if(a[i] != b[i]) {f = 0; break;}
33 }
34 if(f) {
35 printf("Insertion Sort\n");
36 sort(b, b + k + 2);
37 for(int i = 0; i < n; i++){
38 printf("%d", b[i]);
39 if(i != n - 1) printf(" ");
40 }
41 } else {
42 printf("Heap Sort\n");
43 int p = n - 1;
44 while(p > 0 && b[p - 1] <= b[p]) {
45 p--;
46 }
47 swap(b[0], b[p]);
48 solve(0, p - 1);
49 for(int i = 0; i < n; i++){
50 printf("%d", b[i]);
51 if(i != n - 1) printf(" ");
52 }
53 }
54 return 0;
55 }
1099 Build A Binary Search Tree
先排序，中序遍历即为填数的顺序。最后再层次遍历输出。
1 // 1099 Build A Binary Search Tree
2 #include <queue>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 queue <int> q;
9 const int N = 105;
10 int lch[N], rch[N], val[N], a[N], cnt = 0;
11
12 void dfs(int u) {
13 if(lch[u] != -1) dfs(lch[u]);
14 val[u] = a[cnt++];
15 if(rch[u] != -1) dfs(rch[u]);
16 }
17
18 int main() {
19 int n, f = 0;
20 scanf("%d", &n);
21 for(int i = 0; i < n; i++) {
22 int l, r;
23 scanf("%d %d", &l, &r);
24 lch[i] = l; rch[i] = r;
25 }
26 for(int i = 0; i < n; i++) scanf("%d", &a[i]);
27 sort(a, a + n);
28 dfs(0);
29 q.push(0);
30 while(!q.empty()) {
31 int cur = q.front();
32 q.pop();
33 if(!f) f = 1;
34 else printf(" ");
35 printf("%d", val[cur]);
36 if(lch[cur] != -1) q.push(lch[cur]);
37 if(rch[cur] != -1) q.push(rch[cur]);
38 }
39 return 0;
40 }
1100 Mars Numbers
模拟。
1 // 1100 Mars Numbers
2 #include <iostream>
3 using namespace std;
4
5 string s;
6 string str[25] = {
7 "tret", "jan", "feb", "mar", "apr", "may", "jun",
8 "jly", "aug", "sep", "oct", "nov", "dec", "tam", "hel",
9 "maa", "huh", "tou", "kes", "hei", "elo", "syy", "lok",
10 "mer", "jou"
11 };
12
13 void cal1() {
14 if(s[0] == '0') {
15 cout << "tret" << endl;
16 return ;
17 }
18 int num = 0;
19 for(int i = 0; i < s.size(); i++) num = 10 * num + (s[i] - '0');
20 if(num / 13) cout << str[num / 13 + 12];
21 if((num / 13) && (num % 13)) cout << " ";
22 if(num % 13) cout << str[num % 13];
23 cout << endl;
24 }
25
26 void cal2() {
27 int t1 = 0, t2 = 0;
28 string s1 = s.substr(0, 3), s2 = "";
29 if(s.size() > 4) s2 = s.substr(4, 3);
30 for(int i = 1; i <= 24; i++) {
31 if(str[i] == s1) t1 = i;
32 if(str[i] == s2) t2 = i;
33 }
34 if(t1 > 12) cout << (t1 - 12) * 13 + t2 << endl;
35 else cout << t1 + t2 << endl;
36 }
37
38 int main() {
39 int n;
40 cin >> n;
41 getchar();
42 for(int i = 1; i <= n; i++) {
43 getline(cin, s);
44 if(s[0] >= '0' && s[0] <= '9') cal1();
45 else cal2();
46 }
47 return 0;
48 }
1101 Quick Sort
用个数组记录下从前往后，对于当前位置的最大值；从后往前，对于当前位置的最小值。最后再遍历一遍数组即可。
1 // 1101 Quick Sort
2 #include <vector>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 vector <int> ans;
9 const int N = 1e5 + 10;
10 int a[N], pre[N], nxt[N];
11
12 int main() {
13 int n;
14 scanf("%d", &n);
15 for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
16 for(int i = 1; i <= n; i++) {
17 pre[i] = max(pre[i - 1], a[i]);
18 }
19 nxt[n + 1] = 1e9 + 10;
20 for(int i = n; i >= 1; i--) {
21 nxt[i] = min(nxt[i + 1], a[i]);
22 }
23 for(int i = 1; i <= n; i++) {
24 if(pre[i] <= a[i] && nxt[i] >= a[i]) {
25 ans.push_back(a[i]);
26 }
27 }
28 printf("%d\n", ans.size());
29 sort(ans.begin(), ans.end());
30 for(int i = 0; i < ans.size(); i++) {
31 if(i == 0) printf("%d", ans[i]);
32 else printf(" %d", ans[i]);
33 }
34 printf("\n");
35 return 0;
36 }
1102 Invert a Binary Tree
先找下根，中序和层序，把先左后右的顺序调换下。
1 // 1102 Invert a Binary Tree
2 #include <queue>
3 #include <cstdio>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 15;
10 int n, root, f;
11 int lch[N], rch[N], fa[N];
12
13 void level_order() {
14 queue <int> q;
15 q.push(root);
16 while(!q.empty()) {
17 int u = q.front();
18 if(!f) f = 1;
19 else printf(" ");
20 printf("%d", u);
21 q.pop();
22 if(rch[u] != -1) q.push(rch[u]);
23 if(lch[u] != -1) q.push(lch[u]);
24 }
25 printf("\n");
26 }
27
28 void in_order(int u) {
29 if(rch[u] != -1) in_order(rch[u]);
30 if(!f) f = 1;
31 else printf(" ");
32 printf("%d", u);
33 if(lch[u] != -1) in_order(lch[u]);
34 }
35
36 int main() {
37 cin >> n;
38 memset(fa, -1, sizeof(fa));
39 memset(lch, -1, sizeof(lch));
40 memset(rch, -1, sizeof(rch));
41 for(int i = 0; i < n; i++) {
42 char c1, c2;
43 cin >> c1 >> c2;
44 if(c1 != '-') {
45 lch[i] = (c1 - '0');
46 fa[(c1 - '0')] = i;
47 }
48 if(c2 != '-') {
49 rch[i] = (c2 - '0');
50 fa[(c2 - '0')] = i;
51 }
52 }
53 for(int i = 0; i < n; i++){
54 if(fa[i] == -1) root = i;
55 }
56 f = 0;
57 level_order();
58 f = 0;
59 in_order(root);
60 printf("\n");
61 return 0;
62 }
1103 Integer Factorization
预处理下，再DFS+剪枝。更新下答案。
1 // 1103 Integer Factorization
2 #include <cstdio>
3 #include <iostream>
4 #include <algorithm>
5 using namespace std;
6
7 int a[405], b[405], ans[405];
8 int n, m, k, c;
9
10 void dfs(int id, int cnt, int val) {
11 b[cnt] = id;
12 if(cnt > m || val > n) return ;
13 if(val == n && cnt == m) {
14 bool f = 0;
15 for(int i = 1; i <= m; i++) {
16 if(ans[i] != b[i]) {
17 if(b[i] > ans[i]) f = 1;
18 break;
19 }
20 }
21 for(int i = 1; i <= m; i++) ans[i] = b[i];
22 return ;
23 }
24 for(int i = id; i <= c && val + a[i] <= n; i++) {
25 dfs(i, cnt + 1, val + a[i]);
26 }
27 }
28
29 int POW(int u, int v) {
30 int res = 1;
31 for(int i = 1; i <= v; i++) res *= u;
32 return res;
33 }
34
35 int main() {
36 scanf("%d%d%d", &n, &m, &k);
37 for(int i = 1; POW(i, k) <= n; i++) {
38 a[i] = POW(i, k);
39 c = i;
40 }
41 for(int i = 1; i <= c; i++) dfs(i, 1, a[i]);
42 int res = 0;
43 for(int i = 1; i <= m; i++) res += a[ans[i]];
44 if(res == n) {
45 printf("%d = %d^%d", n, ans[m], k);
46 for(int i = m - 1; i >= 1; i--) {
47 printf(" + %d^%d", ans[i], k);
48 }
49 printf("\n");
50 } else {
51 printf("Impossible\n");
52 }
53 return 0;
54 }
1104 Sum of Number Segments
每个位置会算到的次数为$i * (n - i + 1)$。
1 #include <cstdio>
2 using namespace std;
3
4 const int N = 1e5 + 10;
5 double a[N];
6
7 int main() {
8 int n;
9 double ans = 0;
10 scanf("%d", &n);
11 for(int i = 1; i <= n; i++) scanf("%lf", &a[i]);
12 for(int i = 1; i <= n; i++) {
13 ans += a[i] * i * (n - i + 1);
14 }
15 printf("%.2f\n", ans);
16 return 0;
17 }
1105 Spiral Matrix
蛇形填数，注意下细节。
1 // 1105 Spiral Matrix
2 #include <cmath>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e4 + 10;
9 int a[N];
10 int ans[N][N];
11
12 int main() {
13 int n, k;
14 scanf("%d", &n);
15 for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
16 if(n == 1) {
17 printf("%d", a[1]);
18 return 0;
19 }
20 sort(a + 1, a + 1 + n);
21 int m = sqrt(1.0 * n);
22 while(n % m != 0) m++;
23 if((n / m) > m) m = n/ m;
24 //m n/m
25 k = n; n = n / m;
26 int i = 1, j = 1;
27 int U = 1, D = m, L = 1, R = n;
28 while(k > 0) {
29 while(k > 0 && j < R) {
30 ans[i][j] = a[k--];
31 j++;
32 }
33 while(k > 0 && i < D) {
34 ans[i][j] = a[k--];
35 i++;
36 }
37 while(k > 0 && j > U) {
38 ans[i][j] = a[k--];
39 j--;
40 }
41 while(k > 0 && i > L) {
42 ans[i][j] = a[k--];
43 i--;
44 }
45 U++; D--; L++; R--;
46 i++; j++;
47 if(k == 1) {
48 ans[i][j] = a[k--];
49 }
50 }
51 for(int i = 1; i <= m; i++) {
52 for(int j = 1; j <= n; j++) {
53 printf("%d%c", ans[i][j], (j == n) ? '\n' : ' ');
54 }
55 }
56 return 0;
57 }
1106 Lowest Price in Supply Chain
跑下树。在叶子节点更新下最小值，最后再遍历下叶子节点即可。
1 // 1106 Lowest Price in Supply Chain
2 #include <cstdio>
3 #include <vector>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 1e5 + 10;
9 int n, k, id;
10 double p, r, val[N], mi = 1e12;
11 vector <int> E[N];
12
13 void dfs(int u) {
14 if(E[u].size() == 0) mi = min(mi, val[u]);
15 for(int j = 0; j < E[u].size(); j++) {
16 int v = E[u][j];
17 val[v] = val[u] * (1.0 + r / 100.0);
18 dfs(v);
19 }
20 }
21
22 int main() {
23 scanf("%d %lf %lf", &n, &p, &r);
24 for(int i = 0; i < n; i++) {
25 scanf("%d", &k);
26 for(int j = 1; j <= k; j++) {
27 scanf("%d", &id);
28 E[i].push_back(id);
29 }
30 }
31 val[0] = p;
32 dfs(0);
33 int cnt = 0;
34 for(int i = 0; i < n; i++) if(val[i] == mi && E[i].size() == 0) cnt++;
35 printf("%.4f %d\n", mi, cnt);
36 return 0;
37 }
1107 Social Clusters
并查集搞搞。
1 // 1107 Social Clusters
2 #include <cstdio>
3 #include <vector>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 1e3 + 10;
10 vector <int> E[N], ans;
11 int fa[N], C[N];
12
13 int fi(int x) {
14 return x == fa[x] ? x : fa[x] = fi(fa[x]);
15 }
16
17 void Union(int x, int y) {
18 int fx = fi(x), fy = fi(y);
19 if(fx != fy) {
20 fa[fy] = fx;
21 }
22 }
23
24 int main() {
25 bool f = 0;
26 int n, k, u, cnt = 0;
27 scanf("%d", &n);
28 for(int i = 1; i <= n; i++) {
29 fa[i] = i;
30 scanf("%d:", &k);
31 while(k--) {
32 scanf("%d", &u);
33 E[u].push_back(i);
34 }
35 }
36 for(int i = 1; i < N; i++) {
37 for(int j = 1; j < E[i].size(); j++) {
38 u = E[i][j];
39 Union(u, E[i][0]);
40 }
41 }
42 for(int i = 1; i <= n; i++) {
43 C[fi(i)]++;
44 if(i == fa[i]) cnt++;
45 }
46 printf("%d\n", cnt);
47 for(int i = 1; i <= n; i++) {
48 if(C[i] != 0) ans.push_back(C[i]);
49 }
50 sort(ans.begin(), ans.end());
51 for(int i = ans.size() - 1; i >= 0; i--) {
52 if(i != ans.size() - 1) printf(" ");
53 printf("%d", ans[i]);
54 }
55 return 0;
56 }
1108 Finding Average
注意 一个合法数字 的输出格式。
1 // 1108 Finding Average
2 #include <cstdio>
3 #include <cstring>
4 using namespace std;
5
6 char str[50];
7
8 double check() {
9 int i = 0, f = 1, cnt = 0;
10 if(str[0] == '-') i = 1, f = -1;
11 int m = strlen(str);
12 for(; i < m; i++) {
13 if(str[i] >= '0' && str[i] <= '9') continue;
14 if(str[i] == '.') {
15 cnt++;
16 if(cnt >= 2) return 2000;
17 }
18 else return 2000;
19 }
20 double res = 0;
21 i = 0;
22 if(str[0] == '-') i = 1, f = -1;
23 for(; i < m; i++) {
24 if(str[i] == '.') {
25 if(i + 3 < m) return 2000;
26 if(i + 2 < m) res += (str[i + 2] - '0') / 100.0;
27 if(i + 1 < m) res += (str[i + 1] - '0') / 10.0;
28 return f * res;
29 }
30 res = 10 * res + 1.0 * (str[i] - '0');
31 }
32 return f * res;
33 }
34
35 int main() {
36 double average = 0;
37 int n, c = 0;
38 scanf("%d", &n);
39 while(n--) {
40 scanf("%s", str);
41 double val = check();
42 if(val >= -1000 && val <= 1000) average += check(), c++;
43 else{
44 printf("ERROR: %s is not a legal number\n", str);
45 }
46 }
47 if(c == 1) printf("The average of %d number is %.2f\n", c, average / c);
48 else if(c > 1) printf("The average of %d numbers is %.2f\n", c, average / c);
49 else printf("The average of 0 numbers is Undefined\n");
50 return 0;
51 }
1109 Group Photo
模拟。
1 // 1109 Group Photo
2 #include <vector>
3 #include <cstdio>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 struct node {
9 int h;
10 string name;
11 } tmp;
12
13 bool cmp(node x, node y) {
14 if(x.h == y.h) return x.name < y.name;
15 return x.h > y.h;
16 }
17
18 vector<node> p;
19
20 int main() {
21 int n, k, per, last, id = 0;
22 cin >> n >> k;
23 per = n / k;
24 last = per + (n % k);
25 for (int i = 1; i <= n; i++) {
26 cin >> tmp.name >> tmp.h;
27 p.push_back(tmp);
28 }
29 sort(p.begin(), p.end(), cmp);
30 for (int i = 0; i < k; i++) {
31 int now = per, f = -1;
32 if(i == 0) now = last;
33 int pos = now / 2;
34 vector<string> ans(n);
35 while(pos >= 0 && pos < now) {
36 ans[pos] = p[id++].name;
37 pos += f;
38 if(f < 0) {
39 f *= -1;
40 f++;
41 } else {
42 f *= -1;
43 f--;
44 }
45 }
46 for (int j = 0; j < now; j++) {
47 cout << ans[j];
48 if(j != now - 1) cout << " ";
49 else cout << endl;
50 }
51 ans.clear();
52 }
53 return 0;
54 }
1110 Complete Binary Tree
完全二叉树递归出的最大下标值一定等于总节点数。
1 // 1110 Complete Binary Tree
2 #include <queue>
3 #include <iostream>
4 using namespace std;
5
6 int fa[25], lch[25], rch[25];
7 int n, root, last, maxindex = 0;
8
9 void dfs(int u, int index) {
10 if(u == -1) return ;
11 if(index > maxindex) {
12 maxindex = index;
13 last = u;
14 }
15 dfs(lch[u], 2 * index);
16 dfs(rch[u], 2 * index + 1);
17 }
18
19 int main() {
20 cin >> n;
21 for(int i = 0; i < n; i++) fa[i] = lch[i] = rch[i] = -1;
22 for(int i = 0; i < n; i++) {
23 string s1, s2;
24 cin >> s1 >> s2;
25 if(s1 != "-") {
26 fa[atoi(s1.c_str())] = i;
27 lch[i] = atoi(s1.c_str());
28 }
29 if(s2 != "-") {
30 fa[atoi(s2.c_str())] = i;
31 rch[i] = atoi(s2.c_str());
32 }
33 }
34 for(int i = 0; i < n; i++) {
35 if(fa[i] == -1) root = i;
36 }
37 dfs(root, 1);
38 if(maxindex == n) {
39 cout << "YES " << last << endl;
40 } else {
41 cout << "NO " << root << endl;
42 }
43 return 0;
44 }
1112 Stucked Keyboard
模拟。
1 // 1112 Stucked Keyboard
2 #include <map>
3 #include <iostream>
4 using namespace std;
5
6 string str;
7 map<char, int> ok;
8
9 int main() {
10 int n, k, c = 1;
11 cin >> k >> str;
12 n = str.size();
13 for (int i = 1; i < n; i++) {
14 if(str[i] == str[i - 1]) {
15 c++;
16 } else {
17 if(c % k != 0) ok[str[i - 1]] = 1;
18 c = 1;
19 }
20 }
21 if(c > 0 && c % k != 0) ok[str[n - 1]] = 1;
22 for (int i = 0; i < n; i++) {
23 if(ok[str[i]] == 0) {
24 cout << str[i];
25 ok[str[i]] = -1;
26 }
27 }
28 cout << endl;
29 c = 1;
30 for (int i = 1; i < n; i++) {
31 if(str[i] == str[i - 1]) {
32 c++;
33 } else {
34 if(c % k == 0) {
35 if(ok[str[i - 1]] != 1) for (int j = 1; j <= c / k; j++) cout << str[i - 1];
36 else for (int j = 1; j <= c; j++) cout << str[i - 1];
37 }
38 else {
39 for (int j = 1; j <= c; j++) cout << str[i - 1];
40 }
41 c = 1;
42 }
43 }
44 if(c > 0 && c % k == 0) {
45 if(ok[str[n - 1]] != 1) for (int j = 1; j <= c / k; j++) cout << str[n - 1];
46 else for (int j = 1; j <= c; j++) cout << str[n - 1];
47 }
48 else {
49 for (int j = 1; j <= c; j++) cout << str[n - 1];
50 }
51 return 0;
52 }
1113 Integer Set Partition
先排序，再前面加一加，后面加一加。
1 // 1113 Integer Set Partition
2 #include <iostream>
3 #include <algorithm>
4 using namespace std;
5
6 typedef long long ll;
7 const int N = 1e5 + 10;
8 ll a[N];
9
10 int main() {
11 int n, m;
12 ll s1 = 0, s2 = 0;
13 cin >> n;
14 for (int i = 1; i <= n; i++) cin >> a[i];
15 sort(a + 1, a + 1 + n);
16 for (int i = n / 2 + 1; i <= n; i++) s1 += a[i];
17 for (int i = 1; i < n / 2 + 1; i++) s2 += a[i];
18 cout << (n % 2) << " ";
19 cout << abs(s1 - s2) << endl;
20 return 0;
21 }
1114 Family Property
暴力模拟即可。
1 // 1114 Family Property
2 #include <map>
3 #include <vector>
4 #include <iostream>
5 #include <algorithm>
6 using namespace std;
7
8 const int N = 10000 + 10;
9 int fa[N], M[N], A[N];
10 int fi(int x) {
11 return fa[x] == x ? x : fa[x] = fi(fa[x]);
12 }
13
14 void Union(int x, int y) {
15 if(x == -1 || y == -1) return ;
16 int fx = fi(x), fy = fi(y);
17 if(fx != fy) {
18 fa[fx] = fy;
19 }
20 }
21
22 void init() {
23 for(int i = 0; i < N; i++) fa[i] = i;
24 }
25
26 int cnt = 0;
27 vector<int> V, E[N];
28 map<int, int> mp;
29
30 struct node {
31 int ansID, ansM;
32 double avg1, avg2;
33 }ans[N];
34
35 map<int, int> vis;
36
37 bool cmp(node x, node y) {
38 if(x.avg2 == y.avg2) return x.ansID < y.ansID;
39 return x.avg2 > y.avg2;
40 }
41
42 int main() {
43 init();
44 int n;
45 cin >> n;
46 for (int i = 1; i <= n; i++) {
47 int id, father, mother, child, k;
48 cin >> id >> father >> mother >> k;
49 Union(id, father);
50 if(father != -1) V.push_back(father);
51 Union(id, mother);
52 if(mother != -1) V.push_back(mother);
53 while (k--) {
54 cin >> child;
55 if(child != -1) V.push_back(child);
56 Union(id, child);
57 }
58 V.push_back(id);
59 cin >> M[id] >> A[id];
60 }
61 for (int i = 0; i < V.size(); i++) {
62 int u = V[i];
63 if(fa[u] == u && !mp[u]) {
64 cnt++;
65 mp[u] = cnt;
66 }
67 }
68 for (int i = 0; i < V.size(); i++) {
69 int u = V[i];
70 if(!vis[u]) E[mp[fi(u)]].push_back(u), vis[u] = 1;
71 }
72 for (int i = 1; i <= cnt; i++) {
73 int minid = 123456;
74 double sum1 = 0;
75 double sum2 = 0;
76 for(int j = 0; j < E[i].size(); j++) {
77 int v = E[i][j];
78 minid = min(minid, v);
79 sum1 += M[v];
80 sum2 += A[v];
81 }
82 ans[i].ansID = minid;
83 ans[i].ansM = E[i].size();
84 ans[i].avg1 = sum1 / E[i].size();
85 ans[i].avg2 = sum2 / E[i].size();
86 }
87 sort(ans + 1, ans + 1 + n, cmp);
88 cout << cnt << endl;
89 for (int i = 1; i <= cnt; i++) {
90 printf("%04d %d %.3f %.3f\n", ans[i].ansID, ans[i].ansM, ans[i].avg1, ans[i].avg2);
91 }
92 return 0;
93 }
1115 Counting Nodes in a BST
用链表实现二叉搜索树的插入。构建完树后，DFS一遍，在对应深度更新节点数。
1 // 1115 Counting Nodes in a BST
2 #include <iostream>
3 using namespace std;
4
5 struct node {
6 int val;
7 struct node *left, *right;
8 };
9
10 const int N = 1000 + 10;
11 int num[N], maxd = -1;
12
13 node *build(node *root, int v) {
14 if(root == NULL) {
15 root = new node();
16 root -> val = v;
17 root -> left = root -> right = NULL;
18 } else if(v > root->val) {
19 root->right = build(root->right, v);
20 } else {
21 root->left = build(root->left, v);
22 }
23 return root;
24 }
25
26 void dfs(node *root, int dp) {
27 if(root == NULL) {
28 maxd = max(maxd, dp);
29 return ;
30 }
31 num[dp]++;
32 dfs(root->left, dp + 1);
33 dfs(root->right, dp + 1);
34 }
35
36 int main() {
37 int n, k;
38 node *root = NULL;
39 cin >> n;
40 while(n--) {
41 cin >> k;
42 root = build(root, k);
43 }
44 dfs(root, 0);
45 cout << num[maxd - 1] << " + " << num[maxd - 2] << " = " << num[maxd - 1] + num[maxd - 2] << endl;
46 return 0;
47 }
1116 Come on! Let's C
模拟。
1 // 1116 Come on! Let's C
2 #include <map>
3 #include <iostream>
4 using namespace std;
5
6 map<int, int> Rank;
7
8 bool check(int x) {
9 if(x < 2) return false;
10 for(int i = 2; i * i <= x; i++) {
11 if(x % i == 0) return false;
12 }
13 return true;
14 }
15
16 int main() {
17 int n, k;
18 cin >> n;
19 for (int i = 1; i <= n; i++) {
20 cin >> k;
21 Rank[k] = i;
22 }
23 cin >> n;
24 while (n--) {
25 cin >> k;
26 printf("%04d: ", k);
27 if(Rank[k] == -1) cout << "Checked" << endl;
28 else if(Rank[k] == 0) cout << "Are you kidding?" << endl;
29 else if(Rank[k] == 1) cout << "Mystery Award" << endl, Rank[k] = -1;
30 else if(check(Rank[k])) cout << "Minion" << endl, Rank[k] = -1;
31 else cout << "Chocolate" << endl, Rank[k] = -1;
32 }
33 return 0;
34 }
1117 Eddington Number
前缀和
1 // 1117 Eddington Number
2 #include <iostream>
3 #include <algorithm>
4 using namespace std;
5
6 const int N = 1e6 + 10;
7 int d[N];
8
9 int main() {
10 int n, k, ans = 0;
11 cin >> n;
12 for(int i = 1; i <= n; i++) {
13 cin >> k;
14 d[k]++;
15 }
16 for(int i = 1; i < N; i++) d[i] += d[i - 1];
17 for(int i = 1; i < N; i++) {
18 if((n - d[i]) >= i) ans = max(ans, i);
19 }
20 cout << ans << endl;
21 return 0;
22 }
1118 Birds in Forest
并查集
1 // 1118 Birds in Forest
2 #include <cstdio>
3 #include <vector>
4 using namespace std;
5
6 const int N = 1e4 + 10;
7 bool vis[N];
8 int fa[N], cnt = 0;
9 vector <int> birds;
10
11 int fi(int x) {
12 return x == fa[x] ? x : fa[x] = fi(fa[x]);
13 }
14
15 void Union(int x, int y) {
16 int fx = fi(x), fy = fi(y);
17 if(fx != fy) {
18 fa[fx] = fy;
19 }
20 }
21
22 void init() {
23 for (int i = 0; i < N; i++) fa[i] = i;
24 }
25
26 int main() {
27 init();
28 int m, n, k, u, v;
29 scanf("%d", &n);
30 while (n--) {
31 scanf("%d", &k);
32 for (int i = 1; i <= k; i++) {
33 scanf("%d", &v);
34 if(!vis[v]) vis[v] = 1, birds.push_back(v);
35 if(i == 1) u = v;
36 else Union(u, v);
37 }
38 }
39 for (int i = 0; i < birds.size(); i++) {
40 u = birds[i];
41 if(fa[u] == u) cnt++;
42 }
43 printf("%d %d\n", cnt, birds.size());
44 scanf("%d", &m);
45 while (m--) {
46 scanf("%d %d", &u, &v);
47 if(fi(u) == fi(v)) printf("Yes\n");
48 else printf("No\n");
49 }
50 return 0;
51 }
1119 Pre- and Post-order Traversals
通过前序和后序先建棵树。（题目中说明可以随意建棵符合的树）再中序遍历即可。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn=35;
5 int preOrder[maxn], postOrder[maxn];
6 bool isUnique=true;
7
8 struct Node{
9 int left=-1,right=-1;
10 }node[maxn];
11
12 void build(int preL,int preR,int postL,int postR){
13 if(preL>=preR){
14 return;
15 }
16 int fa=preL;
17 int val=preOrder[preL+1];
18 int postIdx;
19 for(int i=postL;i<postR;i++){
20 if(val==postOrder[i]){
21 postIdx=i;
22 break;
23 }
24 }
25 int num=postIdx-postL;
26 if(preR-preL-1==num){
27 isUnique=false;
28 }
29 node[fa].left=preL+1;
30 build(preL+1,preL+num+1,postL,postIdx);
31 if(preR-preL-1>num){
32 node[fa].right=preL+num+2;
33 build(preL+num+2,preR,postIdx+1,postR-1);
34 }
35 }
36
37 bool first=true;
38
39 void inOrder(int root){
40 if(root==-1){
41 return;
42 }
43 inOrder(node[root].left);
44 if(first){
45 first=false;
46 printf("%d",preOrder[root]);
47 }
48 else
49 printf(" %d",preOrder[root]);
50 inOrder(node[root].right);
51 }
52
53 int main(){
54 int n;
55 scanf("%d",&n);
56 for(int i=1;i<=n;i++) scanf("%d",&preOrder[i]);
57 for(int i=1;i<=n;i++) scanf("%d",&postOrder[i]);
58 build(1,n,1,n);
59 if(isUnique) printf("Yes\n");
60 else printf("No\n");
61 inOrder(1);
62 printf("\n");
63 return 0;
64 }
1120 Friend Numbers
模拟
1 // 1120 Friend Numbers
2 #include <set>
3 #include <cstdio>
4 #include <iostream>
5 using namespace std;
6
7 const int N = 1e4 + 10;
8 int a[N], b[N];
9
10 int cal(int x) {
11 int sum1 = 0;
12 while (x) {
13 sum1 += x % 10;
14 x /= 10;
15 }
16 return sum1;
17 }
18
19 int check(int x, int y) {
20 if (x == y) return x;
21 return 0;
22 }
23
24 set<int> ans;
25
26 int main() {
27 int n;
28 bool f = 0;
29 scanf("%d", &n);
30 for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = cal(a[i]);
31 for (int i = 1; i <= n; i++) {
32 for (int j = i; j <= n; j++) {
33 int val = check(b[i], b[j]);
34 if (val) ans.insert(val);
35 }
36 }
37 printf("%d\n", ans.size());
38 for (auto x : ans) {
39 if (!f) f = 1;
40 else printf(" ");
41 printf("%d", x);
42 }
43 return 0;
44 }
1121 Damn Single
模拟
1 // 1121 Damn Single
2 #include <map>
3 #include <vector>
4 #include <cstdio>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 map<int, int> m, vis;
10 vector<int> p;
11 const int N = 1e4 + 10;
12 int a[N];
13
14 int main() {
15 int n, x, y;
16 scanf("%d", &n);
17 while (n--) {
18 scanf("%d %d", &x, &y);
19 x++; y++;
20 m[x] = y;
21 m[y] = x;
22 }
23 scanf("%d", &n);
24 for (int i = 1; i <= n; i++) {
25 scanf("%d", &a[i]);
26 a[i]++;
27 vis[a[i]] = 1;
28 }
29 for (int i = 1; i <= n; i++) {
30 if (vis[ m[a[i]] ]) continue;
31 p.push_back(a[i]);
32 }
33 sort(p.begin(), p.end());
34 printf("%d\n", p.size());
35 for (int i = 0; i < p.size(); i++) {
36 if (i != 0) printf(" ");
37 printf("%05d", p[i] - 1);
38 }
39 return 0;
40 }
1122 Hamiltonian Cycle
模拟
1 // 1122 Hamiltonian Cycle
2 #include <set>
3 #include <cstdio>
4 #include <cstring>
5 #include <iostream>
6 #include <algorithm>
7 using namespace std;
8
9 const int N = 205;
10 int MAP[N][N];
11 set<int> s;
12
13 int main() {
14 memset(MAP, -1, sizeof(MAP));
15 int n, m, k;
16 scanf("%d %d", &n, &m);
17 for (int i = 1; i <= m; i++) {
18 int u, v;
19 scanf("%d %d", &u, &v);
20 MAP[u][v] = 1; MAP[v][u] = 1;
21 }
22 scanf("%d", &k);
23 while (k--) {
24 bool f = 1;
25 int u, v, root;
26 scanf("%d", &m);
27 scanf("%d", &root);
28 s.insert(root);
29 u = root;
30 for (int i = 1; i < m; i++) {
31 scanf("%d", &v);
32 s.insert(v);
33 if (MAP[u][v] == -1) f = 0;
34 u = v;
35 }
36 if (root != u || s.size() != n || m != n + 1) f = 0;
37 if (!f) printf("NO\n");
38 else printf("YES\n");
39 s.clear();
40 }
41 return 0;
42 }
来源：oschina
链接：https://my.oschina.net/u/4273421/blog/3587558