poj-1149-PIGS(最大流)

传送门

这道题最麻烦的就是建图，详细的建图过程在 这里

建完图就直接用dinic算法求最大流就行了

#include <iostream> 
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#define N 1010
const int INF = 0x3f3f3f3f;
using namespace std;
int cap[N][N], level[N], n, m, q[N], house[N], last[N];
bool bfs(int st, int en){
    int rear = 0, front = 0;
    memset(level, 0, sizeof(level));
    level[st] = 1;
    q[front++] = st;
    while(rear < front){
        int u = q[rear++];
        if (u == en)    return true;
        for (int v = 0; v <= m+n+1; v++){
            if (cap[u][v] && level[v] == 0){
                level[v] = level[u] + 1;
                q[front++] = v;
            }
        }
    }
    return false;
}
int dfs(int u, int maxf, int en){
    if (u == en){
        return maxf;
    }
    int ret = 0;
    for (int v = 0; v <= m+n+1; v++){
        if (cap[u][v] && level[v] == level[u]+1){
            int f = dfs(v, min(maxf-ret, cap[u][v]), en);
            ret += f;
            cap[u][v] -= f;
            cap[v][u] += f;
            if (ret == maxf)    return ret;
        }
    }
    return ret;
}
int dinic(int st, int en){
    int ans = 0;
    while(bfs(st, en)){
        ans += dfs(st, INF, en);
    }
    return ans;
}
int main(){
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    int i, j, k, x, y, v;
    while(~scanf("%d%d", &m, &n)){
        memset(cap, 0, sizeof(cap));
        memset(last, 0, sizeof(last));
        for (i = 1; i <= m; i++){
            scanf("%d", &house[i]);
        }
        for (i = 1; i <= n; i++){
            scanf("%d", &j);
            while(j--){
                scanf("%d", &v);
                if (!last[v]){
                    last[v] = i;
                    cap[0][i] += house[v];
                }else{
                    cap[last[v]][i] = INF;
                }
            }
            scanf("%d", &v);
            cap[i][n+1] += v;
        }
        printf("%d\n", dinic(0, n+1));
    }
    return 0;
}