[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 = 511;
const int maxm = 20010;
int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[maxm];
int S, T, N;

void init() {
  memset(dhead, -1, sizeof(dhead));
  for(int i = 0; i < maxn; i++) dedge[i].next = -1;
  cnt = 0;
}

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 = 0; i < n; i++) dd[i] = inf;
  dd[s] = 0;
  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] + 1 && dedge[i].w > 0) {
        dd[dedge[i].v] = dd[u] + 1;
        if(dedge[i].v == t) return 1;
        q.push(dedge[i].v);
      }
    }
  }
  return 0;
}

int dinic(int s, int t, int n) {
  int st[maxn], top;
  int u;
  int flow = 0;
  while(bfs(s, t, n)) {
    for(int i = 0; i < n; i++) cur[i] = dhead[i];
    u = s; top = 0;
    while(cur[s] != -1) {
      if(u == t) {
        int tp = inf;
        for(int i = top - 1; i >= 0; i--) {
          tp = min(tp, dedge[st[i]].w);
        }
        flow += tp;
        for(int i = top - 1; i >= 0; i--) {
          dedge[st[i]].w -= tp;
          dedge[st[i] ^ 1].w += tp;
          if(dedge[st[i]].w == 0) top = i;
        }
        u = dedge[st[top]].u;
      }
      else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
        st[top++] = cur[u];
        u = dedge[cur[u]].v;
      }
      else {
        while(u != s && cur[u] == -1) {
          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 = 0, T = n + m + 1; N = n + m + 2;
    init();
    int ss = 0;
    for(int i = 1; i <= m; i++) {
      scanf("%d", &a);
      ss += a;
      adde(i+n, T, a, 0);
    }
    for(int i = 1; i <= n; i++) {
      scanf("%d %d", &a, &b);
      adde(S, i, a, 0);
      for(int j = 0; j < b; j++) {
        scanf("%d", &v);
        adde(i, n+v, 1, 0);
      }
    }
    int ret = dinic(S, T, N);
    if(ret == ss) puts("Yes");
    else puts("No");
  }
  return 0;
}