题意：给出一张无向图，每个点上都有一个权值，然后让你删掉权值之和尽量小的点，使得S到H不连通。
无向图带权点连通度问题。每个点拆点，容留为点权，对于原来图中的无向边(u,v)，拆成两条边(u',v,inf),(v',u,inf)，跑S'到T 的maxflow就可以了。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define INF 1<<30 6 #define maxn 210 7 #define maxm 30000 8 using namespace std; 9 10 int v[maxm],next[maxm],w[maxm]; 11 int first[maxn],d[maxn],work[maxn],q[maxn]; 12 int e,S,T,n,m; 13 14 void init(){ 15 e = 0; 16 memset(first,-1,sizeof(first)); 17 } 18 19 void add_edge(int a,int b,int c){ 20 v[e] = b;next[e] = first[a];w[e] = c;first[a] = e++; 21 v[e] = a;next[e] = first[b];w[e] = 0;first[b] = e++; 22 } 23 24 int bfs(){ 25 int rear = 0; 26 memset(d,-1,sizeof(d)); 27 d[S] = 0;q[rear++] = S; 28 for(int i = 0;i < rear;i++){ 29 for(int j = first[q[i]];j != -1;j = next[j]) 30 if(w[j] && d[v[j]] == -1){ 31 d[v[j]] = d[q[i]] + 1; 32 q[rear++] = v[j]; 33 if(v[j] == T) return 1; 34 } 35 } 36 return 0; 37 } 38 39 int dfs(int cur,int a){ 40 if(cur == T) return a; 41 for(int &i = work[cur];i != -1;i = next[i]){ 42 if(w[i] && d[v[i]] == d[cur] + 1) 43 if(int t = dfs(v[i],min(a,w[i]))){ 44 w[i] -= t;w[i^1] += t; 45 return t; 46 } 47 } 48 return 0; 49 } 50 51 int dinic(){ 52 int ans = 0; 53 while(bfs()){ 54 memcpy(work,first,sizeof(first)); 55 while(int t = dfs(S,INF)) ans += t; 56 } 57 return ans; 58 } 59 60 int main() 61 { 62 int kase; 63 scanf("%d",&kase); 64 while(kase--){ 65 scanf("%d%d%d%d",&n,&m,&S,&T); 66 S += n; 67 init(); 68 for(int i = 1;i <= n;i++){ 69 int tmp; 70 scanf("%d",&tmp); 71 add_edge(i,i+n,tmp); 72 } 73 for(int i = 1;i <= m;i++){ 74 int a,b; 75 scanf("%d%d",&a,&b); 76 add_edge(a+n,b,INF); 77 add_edge(b+n,a,INF); 78 } 79 printf("%d ",dinic()); 80 } 81 return 0; 82 }