诈个尸。
二分图匹配。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7 const int INF = 1e9;
8 const int maxn = 2e5 + 10;
9 int lv[maxn], it[maxn];
10 int cnt, h[maxn];
11
12 struct edge
13 {
14 int to, pre, cap;
15 } e[maxn<<1];
16
17 void init()
18 {
19 memset(h, -1, sizeof(h));
20 cnt = 0;
21 }
22
23 void add(int from, int to, int cap)
24 {
25 e[cnt].pre = h[from];
26 e[cnt].to = to;
27 e[cnt].cap = cap;
28 h[from] = cnt;
29 cnt++;
30 }
31
32 void ad(int from, int to, int cap)
33 {
34 add(from, to, cap);
35 add(to, from, 0);
36 }
37
38 void bfs(int s)
39 {
40 memset(lv, -1, sizeof(lv));
41 queue<int> q;
42 lv[s] = 0;
43 q.push(s);
44 while(!q.empty())
45 {
46 int v = q.front(); q.pop();
47 for(int i = h[v]; i >= 0; i = e[i].pre)
48 {
49 int cap = e[i].cap, to = e[i].to;
50 if(cap > 0 && lv[to] < 0)
51 {
52 lv[to] = lv[v] + 1;
53 q.push(to);
54 }
55 }
56 }
57 }
58
59 int dfs(int v, int t, int f)
60 {
61 if(v == t) return f;
62 for(int &i = it[v]; i >= 0; i = e[i].pre)
63 {
64 int &cap = e[i].cap, to = e[i].to;
65 if(cap > 0 && lv[v] < lv[to])
66 {
67 int d = dfs(to, t, min(f, cap));
68 if(d > 0)
69 {
70 cap -= d;
71 e[i^1].cap += d;
72 return d;
73 }
74 }
75 }
76 return 0;
77 }
78
79 int Dinic(int s, int t)
80 {
81 int flow = 0;
82 while(1)
83 {
84 bfs(s);
85 if(lv[t] < 0) return flow;
86 memcpy(it, h, sizeof(it));
87 int f;
88 while((f = dfs(s, t, INF)) > 0) flow += f;
89 }
90 }
91
92 int main(void)
93 {
94 init();
95 int m, n;
96 scanf("%d %d", &m, &n);
97 while(1)
98 {
99 int u, v;
100 scanf("%d %d", &u, &v);
101 if(u == -1) break;
102 ad(u, v, 1);
103 }
104 int S = n + 1, T = S + 1;
105 for(int i = 1; i <= m; i++) ad(S, i, 1);
106 for(int i = m + 1; i <= n; i++) ad(i, T, 1);
107 int ans = Dinic(S, T);
108 if(!ans) puts("No Solution!");
109 else
110 {
111 printf("%d\n", ans);
112 for(int i = 1; i <= m; i++)
113 for(int j = h[i]; j != -1; j = e[j].pre)
114 if(e[j].to <= n && !e[j].cap) printf("%d %d\n", i, e[j].to);
115 }
116 return 0;
117 }
2.太空飞行计划问题
最大权闭合图转最大流。
见《最小割模型在信息学竞赛中的应用》。
同时注意割边必满流，满流不一定是割边。
但是Dinic找割点是很方便的，看lv就好。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 #include <vector>
7 using namespace std;
8 const int INF = 1e9;
9 const int maxn = 2e5 + 10;
10 int lv[maxn], it[maxn];
11 int cnt, h[maxn];
12 vector<int> E, I;
13
14 struct edge
15 {
16 int to, pre, cap;
17 } e[maxn<<1];
18
19 void init()
20 {
21 memset(h, -1, sizeof(h));
22 cnt = 0;
23 }
24
25 void add(int from, int to, int cap)
26 {
27 e[cnt].pre = h[from];
28 e[cnt].to = to;
29 e[cnt].cap = cap;
30 h[from] = cnt;
31 cnt++;
32 }
33
34 void ad(int from, int to, int cap)
35 {
36 add(from, to, cap);
37 add(to, from, 0);
38 }
39
40 void bfs(int s)
41 {
42 memset(lv, -1, sizeof(lv));
43 queue<int> q;
44 lv[s] = 0;
45 q.push(s);
46 while(!q.empty())
47 {
48 int v = q.front(); q.pop();
49 for(int i = h[v]; i >= 0; i = e[i].pre)
50 {
51 int cap = e[i].cap, to = e[i].to;
52 if(cap > 0 && lv[to] < 0)
53 {
54 lv[to] = lv[v] + 1;
55 q.push(to);
56 }
57 }
58 }
59 }
60
61 int dfs(int v, int t, int f)
62 {
63 if(v == t) return f;
64 for(int &i = it[v]; i >= 0; i = e[i].pre)
65 {
66 int &cap = e[i].cap, to = e[i].to;
67 if(cap > 0 && lv[v] < lv[to])
68 {
69 int d = dfs(to, t, min(f, cap));
70 if(d > 0)
71 {
72 cap -= d;
73 e[i^1].cap += d;
74 return d;
75 }
76 }
77 }
78 return 0;
79 }
80
81 int Dinic(int s, int t)
82 {
83 int flow = 0;
84 while(1)
85 {
86 bfs(s);
87 if(lv[t] < 0) return flow;
88 memcpy(it, h, sizeof(it));
89 int f;
90 while((f = dfs(s, t, INF)) > 0) flow += f;
91 }
92 }
93
94 int main(void)
95 {
96 init();
97 int m, n, tot = 0;
98 scanf("%d %d", &m, &n);
99 int S = m + n + 1, T = S + 1;
100 for(int i = 1; i <= m; i++)
101 {
102 int p, x;
103 scanf("%d", &p);
104 tot += p;
105 ad(S, i, p);
106 while(getchar() == ' ')
107 {
108 scanf("%d", &x);
109 ad(i, x + m, INF);
110 }
111 }
112 for(int i = 1; i <= n; i++)
113 {
114 int c;
115 scanf("%d", &c);
116 ad(i + m, T, c);
117 }
118 int ans = tot - Dinic(S, T);
119 for(int i = 1; i <= m; i++) if(lv[i] != -1) printf("%d ", i); puts("");
120 for(int i = 1; i <= n; i++) if(lv[i+m] != -1) printf("%d ", i); puts("");
121 printf("%d\n", ans);
122 return 0;
123 }
3.最小路径覆盖问题
DAG上最小路径覆盖转二分图匹配。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7 const int INF = 1e9;
8 const int maxn = 2e5 + 10;
9 int lv[maxn], it[maxn];
10 int cnt, h[maxn];
11 int n, m;
12
13 struct edge
14 {
15 int to, pre, cap;
16 } e[maxn<<1];
17
18 void init()
19 {
20 memset(h, -1, sizeof(h));
21 cnt = 0;
22 }
23
24 void add(int from, int to, int cap)
25 {
26 e[cnt].pre = h[from];
27 e[cnt].to = to;
28 e[cnt].cap = cap;
29 h[from] = cnt;
30 cnt++;
31 }
32
33 void ad(int from, int to, int cap)
34 {
35 add(from, to, cap);
36 add(to, from, 0);
37 }
38
39 void bfs(int s)
40 {
41 memset(lv, -1, sizeof(lv));
42 queue<int> q;
43 lv[s] = 0;
44 q.push(s);
45 while(!q.empty())
46 {
47 int v = q.front(); q.pop();
48 for(int i = h[v]; i >= 0; i = e[i].pre)
49 {
50 int cap = e[i].cap, to = e[i].to;
51 if(cap > 0 && lv[to] < 0)
52 {
53 lv[to] = lv[v] + 1;
54 q.push(to);
55 }
56 }
57 }
58 }
59
60 int dfs(int v, int t, int f)
61 {
62 if(v == t) return f;
63 for(int &i = it[v]; i >= 0; i = e[i].pre)
64 {
65 int &cap = e[i].cap, to = e[i].to;
66 if(cap > 0 && lv[v] < lv[to])
67 {
68 int d = dfs(to, t, min(f, cap));
69 if(d > 0)
70 {
71 cap -= d;
72 e[i^1].cap += d;
73 return d;
74 }
75 }
76 }
77 return 0;
78 }
79
80 int Dinic(int s, int t)
81 {
82 int flow = 0;
83 while(1)
84 {
85 bfs(s);
86 if(lv[t] < 0) return flow;
87 memcpy(it, h, sizeof(it));
88 int f;
89 while((f = dfs(s, t, INF)) > 0) flow += f;
90 }
91 }
92
93 void ans_print(int x)
94 {
95 printf("%d ", x);
96 for(int i = h[x]; i >= 0; i = e[i].pre)
97 if(!e[i].cap && e[i].to <= n + n)
98 ans_print(e[i].to - n);
99 }
100
101 int main(void)
102 {
103 init();
104 scanf("%d %d", &n, &m);
105 int S = n + n + 1, T = S + 1;
106 for(int i = 1; i <= n; i++) ad(S, i, 1), ad(n + i, T, 1);
107 for(int i = 0; i < m; i++)
108 {
109 int u, v;
110 scanf("%d %d", &u, &v);
111 ad(u, v + n, 1);
112 }
113 int ans = n - Dinic(S, T);
114 for(int i = h[T]; i >= 0; i = e[i].pre)
115 {
116 if(e[i].cap) continue;
117 ans_print(e[i].to - n), puts("");
118 }
119 printf("%d\n", ans);
120 return 0;
121 }
4.魔术球问题
转最小路径覆盖问题。（然而上界不是很好估计，大概不是很大。
如果二分，每次要重建图，所以不如枚举快。
枚举每次加边继续Dinic就可以了。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 #include <vector>
7 using namespace std;
8 const int INF = 1e9;
9 const int maxn = 2e4 + 10;
10 const int maxm = 2e5;
11 const int m = 1e4;
12 int lv[maxn], it[maxn];
13 int cnt, h[maxn];
14 vector<int> is;
15
16 struct edge
17 {
18 int to, pre, cap;
19 } e[maxm<<1];
20
21 void init()
22 {
23 memset(h, -1, sizeof(h));
24 cnt = 0;
25 }
26
27 void add(int from, int to, int cap)
28 {
29 e[cnt].pre = h[from];
30 e[cnt].to = to;
31 e[cnt].cap = cap;
32 h[from] = cnt;
33 cnt++;
34 }
35
36 void ad(int from, int to, int cap)
37 {
38 add(from, to, cap);
39 add(to, from, 0);
40 }
41
42 void bfs(int s)
43 {
44 memset(lv, -1, sizeof(lv));
45 queue<int> q;
46 lv[s] = 0;
47 q.push(s);
48 while(!q.empty())
49 {
50 int v = q.front(); q.pop();
51 for(int i = h[v]; i >= 0; i = e[i].pre)
52 {
53 int cap = e[i].cap, to = e[i].to;
54 if(cap > 0 && lv[to] < 0)
55 {
56 lv[to] = lv[v] + 1;
57 q.push(to);
58 }
59 }
60 }
61 }
62
63 int dfs(int v, int t, int f)
64 {
65 if(v == t) return f;
66 for(int &i = it[v]; i >= 0; i = e[i].pre)
67 {
68 int &cap = e[i].cap, to = e[i].to;
69 if(cap > 0 && lv[v] < lv[to])
70 {
71 int d = dfs(to, t, min(f, cap));
72 if(d > 0)
73 {
74 cap -= d;
75 e[i^1].cap += d;
76 return d;
77 }
78 }
79 }
80 return 0;
81 }
82
83 int Dinic(int s, int t)
84 {
85 int flow = 0;
86 while(1)
87 {
88 bfs(s);
89 if(lv[t] < 0) return flow;
90 memcpy(it, h, sizeof(it));
91 int f;
92 while((f = dfs(s, t, INF)) > 0) flow += f;
93 }
94 }
95
96 void ans_print(int x)
97 {
98 printf("%d ", x);
99 for(int i = h[x]; i >= 0; i = e[i].pre)
100 if(!e[i].cap && e[i].to <= m + m)
101 ans_print(e[i].to - m);
102 }
103
104 int main(void)
105 {
106 for(int i = 1; i * i < maxn; i++) is.push_back(i * i);
107 init();
108 int n, ans = 0;
109 scanf("%d", &n);
110 int S = m + m + 1, T = S + 1;
111 for(int i = 1; ; i++)
112 {
113 ad(S, i, 1), ad(i + m, T, 1);
114 for(int j = 0; is[j] < i + i; j++)
115 if(is[j] > i) ad(is[j] - i, i + m, 1);
116 ans += Dinic(S, T);
117 if(i - ans > n) {ans = i - 1; break;}
118 }
119 init();
120 for(int i = 1; i <= ans; i++)
121 {
122 ad(S, i, 1), ad(i + m, T, 1);
123 for(int j = 0; is[j] < i + i; j++)
124 if(is[j] > i) ad(is[j] - i, i + m, 1);
125 }
126 Dinic(S, T);
127 printf("%d\n", ans);
128 for(int i = h[T]; i >= 0; i = e[i].pre)
129 {
130 if(e[i].cap) continue;
131 ans_print(e[i].to - m), puts("");
132 }
133 return 0;
134 }
5.圆桌问题
二分图匹配。多case哇。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7 const int INF = 1e9;
8 const int maxn = 2e5 + 10;
9 int lv[maxn], it[maxn];
10 int cnt, h[maxn];
11
12 struct edge
13 {
14 int to, pre, cap;
15 } e[maxn<<1];
16
17 void init()
18 {
19 memset(h, -1, sizeof(h));
20 cnt = 0;
21 }
22
23 void add(int from, int to, int cap)
24 {
25 e[cnt].pre = h[from];
26 e[cnt].to = to;
27 e[cnt].cap = cap;
28 h[from] = cnt;
29 cnt++;
30 }
31
32 void ad(int from, int to, int cap)
33 {
34 add(from, to, cap);
35 add(to, from, 0);
36 }
37
38 void bfs(int s)
39 {
40 memset(lv, -1, sizeof(lv));
41 queue<int> q;
42 lv[s] = 0;
43 q.push(s);
44 while(!q.empty())
45 {
46 int v = q.front(); q.pop();
47 for(int i = h[v]; i >= 0; i = e[i].pre)
48 {
49 int cap = e[i].cap, to = e[i].to;
50 if(cap > 0 && lv[to] < 0)
51 {
52 lv[to] = lv[v] + 1;
53 q.push(to);
54 }
55 }
56 }
57 }
58
59 int dfs(int v, int t, int f)
60 {
61 if(v == t) return f;
62 for(int &i = it[v]; i >= 0; i = e[i].pre)
63 {
64 int &cap = e[i].cap, to = e[i].to;
65 if(cap > 0 && lv[v] < lv[to])
66 {
67 int d = dfs(to, t, min(f, cap));
68 if(d > 0)
69 {
70 cap -= d;
71 e[i^1].cap += d;
72 return d;
73 }
74 }
75 }
76 return 0;
77 }
78
79 int Dinic(int s, int t)
80 {
81 int flow = 0;
82 while(1)
83 {
84 bfs(s);
85 if(lv[t] < 0) return flow;
86 memcpy(it, h, sizeof(it));
87 int f;
88 while((f = dfs(s, t, INF)) > 0) flow += f;
89 }
90 }
91
92 int main(void)
93 {
94 int N, M;
95 scanf("%d %d", &N, &M);
96 init();
97 int sum = 0;
98 int S = N + M + 1, T = S + 1;
99 for(int i = 1; i <= N; i++)
100 {
101 int cap;
102 scanf("%d", &cap);
103 sum += cap;
104 ad(S, i, cap);
105 }
106 for(int i = 1; i <= M; i++)
107 {
108 int cap;
109 scanf("%d", &cap);
110 ad(i + N, T, cap);
111 }
112 for(int i = 1; i <= N; i++)
113 for(int j = 1; j <= M; j++)
114 ad(i, j + N, 1);
115 if(Dinic(S, T) == sum)
116 {
117 puts("1");
118 for(int i = 1; i <= N; i++)
119 {
120 int fst = 1;
121 for(int j = h[i]; j >= 0; j = e[j].pre)
122 {
123 if(e[j].to != S && e[j].cap == 0)
124 {
125 if(fst) printf("%d", e[j].to - N);
126 else printf(" %d", e[j].to - N);
127 fst = 0;
128 }
129 }
130 puts("");
131 }
132 }
133 else puts("0");
134 return 0;
135 }
第一问随便搞。后面拆点限流跑跑。
1 #include <iostream>
2 #include <cstdio>
3 #include <cstring>
4 #include <algorithm>
5 #include <queue>
6 using namespace std;
7 const int INF = 1e9;
8 const int maxn = 2e5 + 10;
9 int lv[maxn], it[maxn];
10 int cnt, h[maxn];
11
12 struct edge
13 {
14 int to, pre, cap;
15 } e[maxn<<1];
16
17 void init()
18 {
19 memset(h, -1, sizeof(h));
20 cnt = 0;
21 }
22
23 void add(int from, int to, int cap)
24 {
25 e[cnt].pre = h[from];
26 e[cnt].to = to;
27 e[cnt].cap = cap;
28 h[from] = cnt;
29 cnt++;
30 }
31
32 void ad(int from, int to, int cap)
33 {
34 add(from, to, cap);
35 add(to, from, 0);
36 }
37
38 void bfs(int s)
39 {
40 memset(lv, -1, sizeof(lv));
41 queue<int> q;
42 lv[s] = 0;
43 q.push(s);
44 while(!q.empty())
45 {
46 int v = q.front(); q.pop();
47 for(int i = h[v]; i >= 0; i = e[i].pre)
48 {
49 int cap = e[i].cap, to = e[i].to;
50 if(cap > 0 && lv[to] < 0)
51 {
52 lv[to] = lv[v] + 1;
53 q.push(to);
54 }
55 }
56 }
57 }
58
59 int dfs(int v, int t, int f)
60 {
61 if(v == t) return f;
62 for(int &i = it[v]; i >= 0; i = e[i].pre)
63 {
64 int &cap = e[i].cap, to = e[i].to;
65 if(cap > 0 && lv[v] < lv[to])
66 {
67 int d = dfs(to, t, min(f, cap));
68 if(d > 0)
69 {
70 cap -= d;
71 e[i^1].cap += d;
72 return d;
73 }
74 }
75 }
76 return 0;
77 }
78
79 int Dinic(int s, int t)
80 {
81 int flow = 0;
82 while(1)
83 {
84 bfs(s);
85 if(lv[t] < 0) return flow;
86 memcpy(it, h, sizeof(it));
87 int f;
88 while((f = dfs(s, t, INF)) > 0) flow += f;
89 }
90 }
91
92 int a[444], dp[444];
93 int main(void)
94 {
95 int n, s = 0;
96 scanf("%d", &n);
97 int S = 2 * n + 1, T = S + 1;
98 for(int i = 1; i <= n; i++)
99 {
100 scanf("%d", a + i);
101 dp[i] = 1;
102 for(int j = 1; j < i; j++)
103 if(a[i] > a[j]) dp[i] = max(dp[i], dp[j] + 1);
104 s = max(s, dp[i]);
105 }
106
107 printf("%d\n", s);
108
109 init();
110 for(int i = 1; i <= n; i++)
111 {
112 ad(i, i + n, 1);
113
114 if(dp[i] == 1) ad(S, i, 1);
115 if(dp[i] == s) ad(i + n, T, 1);
116
117 for(int j = 1; j <= i; j++)
118 if(a[j] < a[i] && dp[j] + 1 == dp[i]) ad(j + n, i, 1);
119 }
120
121 int ans = Dinic(S, T);
122 printf("%d\n", ans);
123
124 ad(1, 1 + n, INF);
125 ad(n, n + n, INF);
126 ad(S, 1, INF);
127 if(dp[n] == s) ad(n + n, T, INF);
128
129 printf("%d\n", ans + Dinic(S, T));
130
131 return 0;
132 }
来源：https://www.cnblogs.com/Aguin/p/5396674.html