N个人，告诉一系列谁有谁的联系方式。 问题：指定S,T两个人，问除这两个人以外，最少踢掉几个人，可以使得这两个人无法联系的上。给出字典序最小的一种方案。
拆点。因为题目要求的是剔除多少个点。我们要把他转换成剔除多少条边。所以要拆点。
枚举删除每一条边，其余边与原图一致，判断新图的最小割是否比原图减少。减少说明是割集的边。
1 // File Name: 1815.cpp 2 // Author: Missa 3 // Created Time: 2013/4/18 星期四 19:31:30 4 5 #include<iostream> 6 #include<cstdio> 7 #include<cstring> 8 #include<algorithm> 9 #include<cmath> 10 #include<queue> 11 #include<stack> 12 #include<string> 13 #include<vector> 14 #include<cstdlib> 15 #include<map> 16 #include<set> 17 using namespace std; 18 #define CL(x,v) memset(x,v,sizeof(x)); 19 #define R(i,st,en) for(int i=st;i<en;++i) 20 #define LL long long 21 22 const int inf = 0x3f3f3f3f; 23 const int maxn = 410; 24 const int maxm = 160010; 25 struct Edge 26 { 27 int v, c, next;//指向点,流量,下一个点 28 }p[maxm << 1]; 29 int head[maxn], e; 30 int d[maxn], cur[maxn];//层次记录，当前弧优化 31 int n, m, st, en; 32 void init() 33 { 34 e = 0; 35 memset(head, -1, sizeof(head)); 36 } 37 void addEdge(int u, int v, int c) 38 { 39 p[e].v = v; p[e].c = c; 40 p[e].next = head[u];head[u] = e++; 41 p[e].v = u; p[e].c = 0; //无向图时逆边赋值流量cap，有向图时赋值0. 42 p[e].next = head[v];head[v] = e++; 43 } 44 int bfs(int st, int en) 45 { 46 queue <int> q; 47 memset(d, 0, sizeof(d)); 48 d[st] = 1; 49 q.push(st); 50 while (!q.empty()) 51 { 52 int u = q.front(); q.pop(); 53 for (int i = head[u]; i != -1; i = p[i].next) 54 { 55 if (p[i].c > 0 && !d[p[i].v]) 56 { 57 d[p[i].v] = d[u] + 1; 58 q.push(p[i].v); 59 } 60 } 61 } 62 return d[en]; 63 } 64 int dfs(int u, int a)//a表示流量 65 { 66 if (u == en || a == 0) return a; 67 int f, flow = 0; 68 for (int& i = cur[u]; i != -1; i = p[i].next) 69 { 70 if (d[u] + 1 == d[p[i].v] && (f = dfs(p[i].v,min(a,p[i].c))) > 0) 71 { 72 p[i].c -= f; 73 p[i^1].c += f; 74 flow += f; 75 a -= f; 76 if (!a) break; 77 } 78 } 79 return flow; 80 } 81 int dinic(int st, int en) 82 { 83 int res = 0; 84 while (bfs(st, en)) 85 { 86 for (int i = 0; i <= n; ++i) 87 cur[i] = head[i]; 88 res += dfs(st, inf); 89 } 90 return res; 91 } 92 int N; 93 int g[210][210]; 94 bool del[210]; 95 void build() 96 { 97 init(); 98 for (int i = 1; i <= N; ++i) 99 { 100 if (del[i]) continue; 101 for (int j = 1; j <= N; ++j) 102 { 103 if (del[j]) continue; 104 if (i == j) 105 addEdge(i, j + N, 1); 106 else if (g[i][j]) 107 addEdge(i + N, j, inf); 108 } 109 } 110 } 111 112 int main() 113 { 114 while(~scanf("%d%d%d",&N,&st,&en)) 115 { 116 memset(del, 0, sizeof(del)); 117 st += N; 118 n = N << 1; 119 for (int i = 1; i <= N; ++i) 120 for(int j = 1; j <= N; ++j) 121 scanf("%d", &g[i][j]); 122 if(g[st - N][en]) 123 { 124 printf("NO ANSWER!"); 125 continue; 126 } 127 build(); 128 int pre = dinic(st, en); 129 int ans = 0; 130 for (int i = 1; i <= N; ++i) 131 { 132 if (pre == 0) break; 133 if (i == st - N || i == en) continue; 134 del[i] = 1; 135 build(); 136 int tmp = dinic(st, en); 137 if (tmp < pre){ans++; pre = tmp;} 138 else del[i] = 0; 139 } 140 printf("%d\n",ans); 141 for (int i = 1; i <= N; ++i) 142 if(del[i]) 143 printf("%d ",i); 144 printf("\n"); 145 } 146 return 0; 147 }