题意为去掉多少个顶点使图不连通，求顶点连通度问题。拆点，构造图，对于<u,v>可以变成<u2,v1> <v2,u1>容量为无穷，<u1,u2>容量为1.那么求出来的最大流（即最小割）就为所需要删除的顶点个数，需要字典序输出，从小到大枚举顶点，如果不加入当前点，最小割变小了的话 ，说明这个点是肯定要删除的。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 #include <algorithm> 6 using namespace std; 7 #define INF 0x3f3f3f 8 const int N = 415; 9 #define M 160015 10 struct node 11 { 12 int u,v,next; 13 int w; 14 } edge[M<<1]; 15 int head[N],t,vis[N],pp[N],dis[N]; 16 int o[N]; 17 int st,en; 18 int x[N][N],f[N]; 19 void init() 20 { 21 t=0; 22 memset(head,-1,sizeof(head)); 23 } 24 void add(int u,int v,int w) 25 { 26 edge[t].u = u; 27 edge[t].v = v; 28 edge[t].w = w; 29 edge[t].next = head[u]; 30 head[u] = t++; 31 edge[t].u = v; 32 edge[t].v = u; 33 edge[t].w = 0; 34 edge[t].next = head[v]; 35 head[v] = t++; 36 } 37 int bfs() 38 { 39 int i,u; 40 int w; 41 memset(dis,-1,sizeof(dis)); 42 queue<int>q; 43 q.push(st); 44 dis[st] = 0; 45 while(!q.empty()) 46 { 47 u = q.front(); 48 q.pop(); 49 for(i = head[u] ; i != -1 ; i = edge[i].next) 50 { 51 int v = edge[i].v; 52 w = edge[i].w; 53 if(dis[v]<0&&w>0) 54 { 55 dis[v] = dis[u]+1; 56 q.push(v); 57 } 58 } 59 } 60 if(dis[en]>0) return 1; 61 return 0; 62 } 63 int dfs(int u,int te) 64 { 65 int i; 66 int s; 67 if(u==en) return te; 68 for(i = head[u] ; i != -1 ; i = edge[i].next) 69 { 70 int v = edge[i].v; 71 int w = edge[i].w; 72 if(w>0&&dis[v]==dis[u]+1&&(s=dfs(v,min(te,w)))) 73 { 74 edge[i].w-=s; 75 edge[i^1].w+=s; 76 return s; 77 } 78 } 79 dis[u] = -1; 80 return 0; 81 } 82 int dinic() 83 { 84 int flow = 0; 85 int res; 86 while(bfs()) 87 { 88 while(res = dfs(st,INF)) 89 flow+=res; 90 } 91 return flow; 92 } 93 int main() 94 { 95 int n,i,j; 96 while(scanf("%d%d%d",&n,&st,&en)!=EOF) 97 { 98 init(); 99 // memset(x,0,sizeof(x)); 100 memset(f,0,sizeof(f)); 101 st+=n; 102 for(i = 1; i <= n ; i++) 103 { 104 for(j = 1; j <= n; j++) 105 { 106 scanf("%d",&x[i][j]); 107 if(i==j) 108 { 109 add(i,i+n,1); 110 } 111 else if(x[i][j]) 112 { 113 add(i+n,j,INF); 114 } 115 } 116 } 117 if(x[st-n][en]) 118 { 119 puts("NO ANSWER!"); 120 continue; 121 } 122 int ans = dinic(); 123 int cnt = 0; 124 for(i = 1; i <= n ; i++) 125 { 126 if(ans==0) break; 127 if(i==st-n||i==en) continue; 128 f[i] = 1; 129 init(); 130 for(j = 1; j <= n ; j++) 131 { 132 if(f[j]) continue; 133 for(int e = 1; e <= n ; e++) 134 { 135 if(f[e]) continue; 136 if(j==e) 137 add(j,j+n,1); 138 else if(x[j][e]) 139 { 140 add(j+n,e,INF); 141 } 142 } 143 } 144 int ts = dinic(); 145 if(ts<ans) 146 { 147 cnt++; 148 ans = ts; 149 } 150 else f[i] = 0; 151 } 152 cout<<cnt<<endl; 153 for(j = 1; j <= n; j++) 154 if(f[j]) 155 printf("%d ",j); 156 printf(" "); 157 } 158 return 0; 159 }