ACM-ICPC 2018 南京网络赛

 1 /*Huyyt*/
 2 #include<bits/stdc++.h>
 3 using namespace std;
 4 typedef long long ll;
 5 const int mod = 998244353;
 6 const int maxn = 1e3 + 5;
 7 int main()
 8 {
 9         int T;
10         cin >> T;
11         while(T--)
12         {
13                 ll x;
14                 scanf("%lld",&x);
15                 printf("%lld\n",x-1);
16         }
17         return 0;
18 }

//A

#include<bits/stdc++.h>
#define lc(x) (2*x)
#define rc(x) (2*x+1)
#define fi first
#define se second
using namespace std;
typedef long long ll;

const ll mod = 998244353;
const int maxn = 1e5+5;
const int maxm = 1e2+5;
int n,m,k;
vector<int> v[maxm];
int lit[maxn];
int near[maxm];

int main(){
    int t;
    scanf("%d",&t);
    for(int quq=1;quq<=t;quq++){
        scanf("%d%d%d",&n,&m,&k);
        int i,j,a,b;
        for(i=1;i<=m;i++)v[i].clear();
        for(i=1;i<=k;i++){
            scanf("%d%d",&a,&b);
            v[b].push_back(a);
        }
        ll ans=0;
        memset(lit,0,sizeof(lit));
        for(i=1;i<=m;i++){
            memset(near,0,sizeof(near));
            for(j=1;j<=n;j++){
                lit[j]++;
            }
            for(j=0;j<v[i].size();j++)lit[v[i][j]]=0;
            for(j=1;j<=n;j++){
                b = near[0];
                for(a=1;a<=lit[j];a++){
                    ans+=(j-b);
                    b = max(b,near[a]);
                }
                near[lit[j]]=j;
            }
        }
        printf("Case #%d: %lld\n",quq,ans);
    }
}

//B

  1 /*Huyyt*/
  2 #include<bits/stdc++.h>
  3 #define mem(a,b) memset(a,b,sizeof(a))
  4 using namespace std;
  5 typedef long long ll;
  6 const int mod = 998244353;
  7 const int maxn = 2e4 + 5;
  8 int n, m;
  9 int num[maxn];
 10 int ren[205][20];
 11 int sum[205];
 12 int ans[205];
 13 int cur, biggestnum;
 14 int biggestman;
 15 int numcur;
 16 bool flag = true;
 17 void get_ans()
 18 {
 19         for (int i = 1; i <= n; i++)
 20         {
 21                 for (int j = 1; j <= 13; j++)
 22                 {
 23                         ans[i] += ren[i][j] * j;
 24                 }
 25         }
 26         return ;
 27 }
 28 int get_min(int x)
 29 {
 30         for (int i = 3; i <= 13; i++)
 31         {
 32                 if (ren[x][i])
 33                 {
 34                         return i;
 35                 }
 36         }
 37         if (ren[x][1])
 38         {
 39                 return 1;
 40         }
 41         if (ren[x][2])
 42         {
 43                 return 2;
 44         }
 45 }
 46 int get_next(int x)
 47 {
 48         if (x == 2)
 49         {
 50                 return -1;
 51         }
 52         if (x >= 3 && x <= 12)
 53         {
 54                 return (x + 1);
 55         }
 56         if (x == 1)
 57         {
 58                 return 2;
 59         }
 60         if (x == 13)
 61         {
 62                 return 1;
 63         }
 64 }
 65 int change_man(int x)
 66 {
 67         if (x > n)
 68         {
 69                 return 1;
 70         }
 71         return x;
 72 }
 73 void chouka(int x)
 74 {
 75         for (int i = x; i <= n && cur <= m; i++)
 76         {
 77                 ren[i][num[cur++]]++;
 78                 sum[i]++;
 79         }
 80         for (int i = 1; i < x && cur <= m; i++)
 81         {
 82                 ren[i][num[cur++]]++;
 83                 sum[i]++;
 84         }
 85 }
 86 void go(int mannow)
 87 {
 88         if (!flag)
 89         {
 90                 return ;
 91         }
 92         mannow = change_man(mannow);
 93         if (mannow == biggestman)
 94         {
 95                 chouka(biggestman);
 96                 return ;
 97         }
 98         if (biggestnum == 2)
 99         {
100                 chouka(biggestman);
101                 return ;
102         }
103         int aim = get_next(biggestnum);
104         if (aim == -1)
105         {
106                 chouka(biggestman);
107                 return ;
108         }
109         if (ren[mannow][aim])
110         {
111                 ren[mannow][aim]--;
112                 sum[mannow]--;
113                 if (sum[mannow] == 0)
114                 {
115                         flag = false;
116                         get_ans();
117                         return ;
118                 }
119                 biggestman = mannow, biggestnum = aim;
120                 go(mannow + 1);
121         }
122         else
123         {
124                 if (ren[mannow][2])
125                 {
126                         ren[mannow][2]--;
127                         sum[mannow]--;
128                         if (sum[mannow] == 0)
129                         {
130                                 flag = false;
131                                 get_ans();
132                                 return ;
133                         }
134                         biggestman = mannow, biggestnum = 2;
135                         go(mannow + 1);
136                 }
137                 else
138                 {
139                         go(mannow + 1);
140                 }
141         }
142 }
143 int main()
144 {
145         int T;
146         cin >> T;
147         for (int cas = 1; cas <= T; cas++)
148         {
149                 flag = true;
150                 numcur = 1, cur = 1, biggestnum = -1;
151                 mem(ren, 0), mem(ans, 0), mem(sum, 0);
152                 scanf("%d %d", &n, &m);
153                 for (int i = 1; i <= m; i++)
154                 {
155                         scanf("%d", &num[i]);
156                 }
157                 for (int i = 1; i <= n; i++)
158                 {
159                         int add = 0;
160                         for (; cur <= m && add <= 4;)
161                         {
162                                 //cout<<i<<" "<<num[cur]<<endl;
163                                 ren[i][num[cur++]]++;
164                                 sum[i]++;
165                                 add++;
166                         }
167                 }
168                 biggestman = 1;
169                 while (flag)
170                 {
171                         if (sum[biggestman] == 0)
172                         {
173                                 get_ans();
174                                 flag = false;
175                                 continue;
176                         }
177                         biggestnum = get_min(biggestman);
178                         ren[biggestman][biggestnum]--;
179                         sum[biggestman]--;
180                         if (sum[biggestman] == 0)
181                         {
182                                 get_ans();
183                                 flag = false;
184                                 continue;
185                         }
186                         go(biggestman + 1);
187                 }
188                 printf("Case #%d:\n", cas);
189                 for (int i = 1; i <= n; i++)
190                 {
191                         if (ans[i] == 0)
192                         {
193                                 printf("Winner\n");
194                         }
195                         else
196                         {
197                                 printf("%d\n", ans[i]);
198                         }
199                 }
200         }
201         return 0;
202 }

//C

  1 /*
  2  Author: LargeDumpling
  3  Email: LargeDumpling@qq.com
  4  Edit History:
  5     2018-09-01    File created.
  6 */
  7 
  8 #include<iostream>
  9 #include<cstdio>
 10 #include<cstdlib>
 11 #include<cstring>
 12 #include<cmath>
 13 #include<algorithm>
 14 using namespace std;
 15 const int MAXN=1050;
 16 const double eps=1e-10;
 17 int dcmp(const double &x) { if(fabs(x)<eps) return 0; return x<0?-1:1; }
 18 typedef struct Poi Vec;
 19 struct Poi
 20 {
 21     double x,y;
 22     Poi(const double &a=0,const double &b=0):x(a),y(b) { }
 23     Poi operator+(const Poi &P)const { return Poi(x+P.x,y+P.y); }
 24     Poi operator-(const Poi &P)const { return Poi(x-P.x,y-P.y); }
 25     Poi operator*(const double &P)const { return Poi(x*P,y*P); }
 26     Poi operator/(const double &P)const { return Poi(x/P,y/P); }
 27     bool operator<(const Poi &P)const { return !dcmp(x-P.x)?dcmp(y-P.y)<=0:dcmp(x-P.x)<=0; }
 28 }PP[MAXN],NP[MAXN<<1],IP[MAXN];
 29 struct Line
 30 {
 31     Poi Ps; Vec Dir;
 32     double ang;
 33     Line(const Poi &ps=Poi(),const Vec &dir=Vec()):Ps(ps),Dir(dir) { ang=atan2(Dir.y,Dir.x); }
 34     bool operator<(const Line &L1)const { return ang<L1.ang; }
 35 }lines[MAXN],q[MAXN];
 36 double dOt(const Vec &V1,const Vec &V2) { return V1.x*V2.x+V1.y*V2.y; }
 37 double cRoss(const Vec &V1,const Vec &V2) { return V1.x*V2.y-V1.y*V2.x; }
 38 double lEnth(const Vec V) { return sqrt(dOt(V,V)); }
 39 Vec Normal(const Vec &V) { return Vec(-V.y,V.x); }
 40 Poi iNtersect(const Poi &P1,const Vec &V1,const Poi &P2,const Vec &V2)
 41 { return P1+V1*(cRoss(V2,P1-P2)/cRoss(V1,V2)); }
 42 Poi iNtersect(const Line &L1,const Line &L2)
 43 { return iNtersect(L1.Ps,L1.Dir,L2.Ps,L2.Dir); }
 44 int HPI(Line *L,int N,Poi *Pol)
 45 {
 46     int l,r,m=0;
 47     sort(L,L+N);//按极角排序
 48     q[l=r=0]=L[0];
 49     for(int i=1;i<N;i++)
 50     {
 51         while(l<r&&dcmp(cRoss(L[i].Dir,IP[r-1]-L[i].Ps))<=0) r--; //新加入的直线可能是尾部的一些交点失效
 52         while(l<r&&dcmp(cRoss(L[i].Dir,IP[l]-L[i].Ps))<=0) l++; //首部
 53         q[++r]=L[i]; //加入
 54         if(!dcmp(cRoss(q[r].Dir,q[r-1].Dir)))
 55         { //对于平行直线要取靠左的
 56             r--;
 57             if(dcmp(cRoss(q[r+1].Dir,q[r].Ps-q[r+1].Ps))<0) q[r]=q[r+1];
 58         }
 59         if(l<r) IP[r-1]=iNtersect(q[r-1],q[r]); //如果队列中有至少两条线，则取交点
 60     }
 61     while(l<r&&dcmp(cRoss(q[l].Dir,IP[r-1]-q[l].Ps))<=0) r--; //后面一些交点可能实际上是无用的
 62     if(r-l<2) return 0; //如果只有不到两条线，则失败了
 63     IP[r]=iNtersect(q[l],q[r]); //将最后一条线和第一条线交起来
 64     for(int i=l;i<=r;i++) Pol[m++]=IP[i];
 65     Pol[m]=Pol[0];
 66     return m;
 67 }
 68 double area(const Poi &P1,const Poi &P2,const Poi &P3)
 69 {
 70     return fabs(cRoss(P3-P1,P2-P1));
 71 }
 72 double calc(int n,Poi *P)
 73 {
 74     int l,r,mid1,mid2,tem;
 75     double ans=0;
 76     for(int i=0;i<=n;i++)
 77         P[i+n]=P[i];
 78     for(int i=0;i<n;i++)
 79         for(int j=i+1;j<n;j++)
 80         {
 81             l=i; r=j;
 82             while(l<r-1)
 83             {
 84                 tem=(r-l+1)/3;
 85                 mid1=l+tem;
 86                 mid2=r-tem;
 87                 if(area(P[i],P[j],P[mid1])<area(P[i],P[j],P[mid2])) l=mid1;
 88                 else r=mid2;
 89             }
 90             ans=max(ans,max(area(P[i],P[j],P[l]),area(P[i],P[j],P[r])));
 91             l=j; r=i+n;
 92             while(l<r-1)
 93             {
 94                 tem=(r-l+1)/3;
 95                 mid1=l+tem;
 96                 mid2=r-tem;
 97                 if(area(P[i],P[j],P[mid1])<area(P[i],P[j],P[mid2])) l=mid1;
 98                 else r=mid2;
 99             }
100             ans=max(ans,max(area(P[j],P[i+n],P[l]),area(P[j],P[i+n],P[r])));
101         }
102     return ans;
103 }
104 int main()
105 {
106     int T_T,n,m;
107     double r;
108     Vec Nv;
109     scanf("%d",&T_T);
110     while(T_T--)
111     {
112         scanf("%d%lf",&n,&r);
113         for(int i=0;i<n;i++) scanf("%lf%lf",&PP[n-i-1].x,&PP[n-i-1].y);
114         PP[n]=PP[0];
115         for(int i=0;i<n;i++)
116         {
117             Nv=Normal(PP[i+1]-PP[i]);
118             Nv=Nv/lEnth(Nv);
119             lines[i]=Line(PP[i]+Nv*r,PP[i+1]-PP[i]);
120         }
121         m=HPI(lines,n,NP);
122         printf("%lf\n",calc(m,NP));
123     }
124     fclose(stdin);
125     fclose(stdout);
126     return 0;
127 }

//D

 1 /*
 2  Author: LargeDumpling
 3  Email: LargeDumpling@qq.com
 4  Edit History:
 5     2018-09-01    File created.
 6 */
 7 
 8 #include<iostream>
 9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<queue>
14 #include<algorithm>
15 using namespace std;
16 const int MAXN=25;
17 struct jz
18 {
19     int u,S,cnt;
20     jz(const int &U=0,const int &SS=0,const int &C=0):u(U),S(SS),cnt(C) { }
21 };
22 bool aa;
23 int n,pre[MAXN];
24 long long a[MAXN],b[MAXN],f[1100000],ans=0;
25 bool exist[1100000],vis[1100000];
26 bool bb;
27 void SPFA()
28 {
29     int S=0;
30     long long cnt=0;
31     queue<jz> q;
32     f[0]=0;
33     for(int i=0;i<n;i++) if((pre[i]&S)==pre[i]
34             &&(!vis[S|(1<<i)]||f[S|(1<<i)]<f[S]+a[i]*(cnt+1LL)+b[i]))
35     {
36         f[S|(1<<i)]=f[S]+a[i]*(cnt+1LL)+b[i];
37         vis[S|(1<<i)]=true;
38         q.push(jz(i,S|(1<<i),cnt+1));
39         exist[S|(1<<i)]=true;
40     }
41     while(q.size())
42     {
43         S=q.front().S;
44         cnt=q.front().cnt;
45         q.pop();
46         exist[S]=false;
47         for(int i=0;i<n;i++) if((!((S>>i)&1))&&((pre[i]&S)==pre[i])
48                 &&(!vis[S|(1<<i)]||f[S|(1<<i)]<f[S]+a[i]*(cnt+1LL)+b[i]))
49         {
50             f[S|(1<<i)]=f[S]+a[i]*(cnt+1LL)+b[i];
51             vis[S|(1<<i)]=true;
52             ans=max(ans,f[S|(1<<i)]);
53             if(!exist[S|(1<<i)])
54             {
55                 exist[S|(1<<i)]=true;
56                 q.push(jz(i,S|(1<<i),cnt+1));
57             }
58         }
59     }
60     return;
61 }
62 int main()
63 {
64     int pn,x;
65     scanf("%d",&n);
66     for(int i=0;i<n;i++)
67     {
68         scanf("%lld%lld%d",&a[i],&b[i],&pn);
69         while(pn--)
70         {
71             scanf("%d",&x); x--;
72             pre[i]|=(1<<x);
73         }
74     }
75     SPFA();
76     printf("%lld\n",ans);
77     fclose(stdin);
78     fclose(stdout);
79     return 0;
80 }

//E

  1 /*
  2  Author: LargeDumpling
  3  Email: LargeDumpling@qq.com
  4  Edit History:
  5     2018-09-01    File created.
  6 */
  7 
  8 #include<iostream>
  9 #include<cstdio>
 10 #include<cstdlib>
 11 #include<cstring>
 12 #include<cmath>
 13 #include<algorithm>
 14 using namespace std;
 15 const int MAXN=100050;
 16 const int M=131072;
 17 const int INF=2147483647;
 18 int n,m;
 19 int mbl[MAXN],d[M<<1];
 20 int ans[MAXN][2];
 21 void read1n(int &x)
 22 {
 23     char ch;
 24     for(ch=getchar();ch<'0'||'9'<ch;ch=getchar())
 25         if(ch==-1) return;
 26     for(x=0;'0'<=ch&&ch<='9';ch=getchar())
 27         x=(x<<1)+(x<<3)+ch-'0';
 28     return;
 29 }
 30 void build()
 31 {
 32     for(int i=1;i<=n;i++)
 33         d[i+M]=mbl[i];
 34     for(int i=M-1;i;i--)
 35         d[i]=min(d[i<<1],d[i<<1|1]);
 36     return;
 37 }
 38 void del(int x)
 39 {
 40     mbl[x]=d[x+M]=INF;
 41     for(int i=(x+M)>>1;i;i>>=1)
 42         d[i]=min(d[i<<1],d[i<<1|1]);
 43     return;
 44 }
 45 int query(int L,int R)
 46 {
 47     int ans=INF;
 48     for(L=L+M-1,R=R+M+1;L<R-1;L>>=1,R>>=1)
 49     {
 50         if(!(L&1)) ans=min(ans,d[L^1]);
 51         if(R&1) ans=min(ans,d[R^1]);
 52     }
 53     return ans;
 54 }
 55 void calc()
 56 {
 57     int nex,cnt,now,l,mid,r,left=0;
 58     ans[0][0]=ans[0][1]=0;
 59     build();
 60     cnt=0;
 61     for(int i=1;i<MAXN;i++)
 62     {
 63         if(cnt==n)
 64         {
 65             ans[i][0]=ans[i-1][0];
 66             ans[i][1]=ans[i-1][1];
 67             continue;
 68         }
 69         left+=m;
 70         now=1;
 71         while(true)
 72         {
 73             if(mbl[now]<=left) nex=now;
 74             else if(left<query(now,n)) nex=n+1;
 75             else
 76             {
 77                 l=now; r=n;
 78                 while(l<r-1)
 79                 {
 80                     mid=(l+r)>>1;
 81                     if(query(now,mid)<=left) r=mid;
 82                     else l=mid;
 83                 }
 84                 nex=r;
 85             }
 86             if(nex==n+1) break;
 87             left-=mbl[nex];
 88             cnt++;
 89             del(nex);
 90             now=nex;
 91         }
 92         ans[i][0]=cnt;
 93         ans[i][1]=left;
 94     }
 95 }
 96 int main()
 97 {
 98     int Q,x;
 99     read1n(n); read1n(m);
100     for(int i=1;i<=n;i++)
101         scanf("%d",&mbl[i]);
102     calc();
103     read1n(Q);
104     while(Q--)
105     {
106         read1n(x);
107         printf("%d %d\n",ans[x][0],ans[x][1]);
108     }
109     fclose(stdin);
110     fclose(stdout);
111     return 0;
112 }

//G

 1 /*
 2  Author: LargeDumpling
 3  Email: LargeDumpling@qq.com
 4  Edit History:
 5     2018-09-01    File created.
 6 */
 7 
 8 #include<iostream>
 9 #include<cstdio>
10 #include<cstdlib>
11 #include<cstring>
12 #include<cmath>
13 #include<algorithm>
14 using namespace std;
15 const int MAXN=2000050;
16 const int MAXC=10;
17 const long long mod=1000000007LL;
18 char str[MAXN];
19 int ch[MAXN][MAXC],len[MAXN],fail[MAXN],last,sz;
20 long long num[MAXN],ans,p[MAXN];
21 int getfail(char T[],int x,int i)
22 {
23     while(T[i-len[x]-1]!=T[i]) x=fail[x];
24     return x;
25 }
26 void init()
27 {
28     memset(ch[0],last=0,sizeof(ch[0]));//last为当前的最长后缀回文
29     memset(ch[1],0,sizeof(ch[1]));
30     len[0]=0; len[1]=-1;
31     fail[0]=1;
32     num[0]=num[1]=0;
33     sz=1;
34     ans=0;
35     return;
36 }
37 void insert(char T[])
38 {
39     int lenth=strlen(T),cur=0;
40     for(int i=0;i<lenth;i++)
41     {
42         cur=getfail(T,last,i);
43         if(!ch[cur][T[i]-'0'])
44         {
45             fail[++sz]=ch[ getfail(T,fail[cur],i) ][T[i]-'0'];
46             ch[cur][T[i]-'0']=sz;
47             memset(ch[sz],0,sizeof(ch[sz]));
48             len[sz]=len[cur]+2;
49             if(cur==1) num[sz]=T[i]-'0';
50             else num[sz]=(num[cur]*10LL%mod+p[len[cur]+1]*(T[i]-'0')%mod+(T[i]-'0'))%mod;
51             ans=(ans+num[sz])%mod;
52         }
53         last=ch[cur][T[i]-'0'];
54     }
55     return;
56 }
57 void pre_calc()
58 {
59     p[0]=1;
60     p[1]=10LL;
61     for(int i=2;i<MAXN;i++) p[i]=p[i-1]*p[1]%mod;
62     return;
63 }
64 int main()
65 {
66     init();
67     pre_calc();
68     scanf("%s",str);
69     insert(str);
70     printf("%lld\n",ans);
71     fclose(stdin);
72     fclose(stdout);
73     return 0;
74 }

//I

 1 #include<bits/stdc++.h>
 2 #define lc(x) (2*x)
 3 #define rc(x) (2*x+1)
 4 #define fi first
 5 #define se second
 6 using namespace std;
 7 typedef long long ll;
 8 
 9 const ll mod = 998244353;
10 const int maxn = 2e7+5;
11 const int maxm = 1e4;
12 int num[maxn],is[maxn];
13 int n;
14 
15 void yuchuli(){
16     int i,j,a,b;
17     for(i=2;;i++){
18         if(i*i>=maxn)break;
19         a=i*i;
20         for(j=a;j<maxn;j+=a)is[j]=1;
21     }
22     for(i=1;i<maxn;i++)num[i]=num[i-1]+1-is[i];
23 }
24 
25 int main(){
26     int t;
27     yuchuli();
28     scanf("%d",&t);
29     while(t--){
30         scanf("%d",&n);
31         int i,j,a,b;
32         ll ans=0;
33         for(i=1;i<=maxm;i++){
34             if(is[i])continue;
35             j=n/i;
36             ans += num[j];
37             if(j>maxm)ans+=num[j]-num[maxm];
38         }
39         printf("%lld\n",ans);
40     }
41 }

//J

  1 /*
  2  Author: LargeDumpling
  3  Email: LargeDumpling@qq.com
  4  Edit History:
  5     2018-09-01    File created.
  6 */
  7 
  8 #include<iostream>
  9 #include<cstdio>
 10 #include<cstdlib>
 11 #include<cstring>
 12 #include<cmath>
 13 #include<queue>
 14 #include<algorithm>
 15 using namespace std;
 16 const int MAXN=100050;
 17 const int MAXM=200050;
 18 struct jz
 19 {
 20     int u;
 21     long long dis;
 22     jz(const int &U=0,const long long &D=0):u(U),dis(D) { }
 23     bool operator<(const jz &X)const { return dis==X.dis?u>X.u:dis>X.dis; }
 24 };
 25 bool aaa;
 26 int T_T,n,m,k;
 27 int fir[MAXN*15],eNd[MAXM*25],nxt[MAXM*25],ed=0;
 28 bool vis[MAXN*15];
 29 long long len[MAXM*25],dis[MAXN*15],ans;
 30 bool bbb;
 31 void addedge(int u,int v,int w)
 32 {
 33     eNd[++ed]=v;
 34     nxt[ed]=fir[u];
 35     len[ed]=w;
 36     fir[u]=ed;
 37     return;
 38 }
 39 void read1n(int &x)
 40 {
 41     char ch=getchar();
 42     while(ch<'0'||'9'<ch) ch=getchar();
 43     for(x=0;'0'<=ch&&ch<='9';ch=getchar())
 44         x=(x<<1)+(x<<3)+ch-'0';
 45     return;
 46 }
 47 void init()
 48 {
 49     memset(fir,ed=0,sizeof(fir));
 50     memset(dis,-1,sizeof(dis));
 51     memset(vis,false,sizeof(vis));
 52     return;
 53 }
 54 void Hijkstra()
 55 {
 56     int u;
 57     priority_queue<jz> q;
 58     dis[1]=0;
 59     q.push(jz(1,0));
 60     while(q.size())
 61     {
 62         while(q.size()&&(vis[q.top().u]||dis[q.top().u]!=q.top().dis))
 63             q.pop();
 64         if(!q.size()) break;
 65         u=q.top().u; q.pop();
 66         vis[u]=true;
 67         for(int i=fir[u];i;i=nxt[i]) if(dis[eNd[i]]==-1||dis[u]+len[i]<dis[eNd[i]])
 68         {
 69             dis[eNd[i]]=dis[u]+len[i];
 70             q.push(jz(eNd[i],dis[eNd[i]]));
 71         }
 72     }
 73     return;
 74 }
 75 int Rand()
 76 {
 77     return (rand()<<15)+rand();
 78 }
 79 int main()
 80 {
 81     int u,v,w;
 82     read1n(T_T);
 83     //T_T=5;
 84     while(T_T--)
 85     {
 86         init();
 87         read1n(n);
 88         //n=100000;
 89         read1n(m);
 90         //m=200000;
 91         read1n(k);
 92         //k=10;
 93         for(int i=1;i<=m;i++)
 94         {
 95             read1n(u);
 96             //u=Rand()%n+1;
 97             read1n(v);
 98             //v=Rand()%n+1;
 99             read1n(w);
100             //w=1000000000;
101             for(int j=0;j<=k;j++)
102             {
103                 addedge(u+j*n,v+j*n,w);
104                 if(j<k) addedge(u+j*n,v+(j+1)*n,0);
105             }
106         }
107         Hijkstra();
108         ans=dis[n];
109         for(int i=1;i<=k;i++)
110             ans=min(ans,dis[(i+1)*n]);
111         printf("%lld\n",ans);
112     }
113     fclose(stdin);
114     fclose(stdout);
115     return 0;
116 }

//L

来源：oschina

链接：https://my.oschina.net/u/4263629/blog/3840227

标签

poi

dcmp

nex