[HIHO1393]网络流三·二分图多重匹配

题目链接：http://hihocoder.com/problemset/problem/1393

把项目到汇点的边权值都加起来，跑完最大流后看是否最大流=权值和。如果等于权值和说明所有项目都有足够的人参与。

 #include <bits/stdc++.h>
 using namespace std;

 typedef struct Edge {
   int u, v, w, next;
 }Edge;
 const int inf = 0x7f7f7f7f;
 const int maxn = ;
 const int maxm = ;
 int cnt, dhead[maxn];
 int cur[maxn], dd[maxn];
 Edge dedge[maxm];
 int S, T, N;

 void init() {
   memset(dhead, -, sizeof(dhead));
   for(int i = ; i < maxn; i++) dedge[i].next = -;
   cnt = ;
 }

 void adde(int u, int v, int w, int c1) {
   dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w;
   dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
   dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1;
   dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
 }

 bool bfs(int s, int t, int n) {
   queue<int> q;
   for(int i = ; i < n; i++) dd[i] = inf;
   dd[s] = ;
   q.push(s);
   while(!q.empty()) {
     int u = q.front(); q.pop();
     for(int i = dhead[u]; ~i; i = dedge[i].next) {
       if(dd[dedge[i].v] > dd[u] +  && dedge[i].w > ) {
         dd[dedge[i].v] = dd[u] + ;
         if(dedge[i].v == t) return ;
         q.push(dedge[i].v);
       }
     }
   }
   return ;
 }

 int dinic(int s, int t, int n) {
   int st[maxn], top;
   int u;
   int flow = ;
   while(bfs(s, t, n)) {
     for(int i = ; i < n; i++) cur[i] = dhead[i];
     u = s; top = ;
     while(cur[s] != -) {
       if(u == t) {
         int tp = inf;
         for(int i = top - ; i >= ; i--) {
           tp = min(tp, dedge[st[i]].w);
         }
         flow += tp;
         for(int i = top - ; i >= ; i--) {
           dedge[st[i]].w -= tp;
           dedge[st[i] ^ ].w += tp;
           if(dedge[st[i]].w == ) top = i;
         }
         u = dedge[st[top]].u;
       }
       else if(cur[u] != - && dedge[cur[u]].w >  && dd[u] +  == dd[dedge[cur[u]].v]) {
         st[top++] = cur[u];
         u = dedge[cur[u]].v;
       }
       else {
         while(u != s && cur[u] == -) {
           u = dedge[st[--top]].u;
         }
         cur[u] = dedge[cur[u]].next;
       }
     }
   }
   return flow;
 }

 int n, m;
 int a, b, v;

 int main() {
   // freopen("in", "r", stdin);
   int Q;
   scanf("%d", &Q);
   while(Q--) {
     scanf("%d%d",&n,&m);
     S = , T = n + m + ; N = n + m + ;
     init();
     int ss = ;
     for(int i = ; i <= m; i++) {
       scanf("%d", &a);
       ss += a;
       adde(i+n, T, a, );
     }
     for(int i = ; i <= n; i++) {
       scanf("%d %d", &a, &b);
       adde(S, i, a, );
       for(int j = ; j < b; j++) {
         scanf("%d", &v);
         adde(i, n+v, , );
       }
     }
     int ret = dinic(S, T, N);
     if(ret == ss) puts("Yes");
     else puts("No");
   }
   return ;
 }