POJ - 3281 -- Dining

题目来源：http://poj.org/problem?id=3281

网络流建图，最大流问题。

边的方向为：s->食物->牛->牛->饮料->t。

将牛拆成两个顶点，之间连一条容量为1的边，这样避免了一头牛杯分配多组食物和饮料的方案，计算出图中的最大流即可。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
struct edge{
    int to,next,cap,rev;
}e[100100];
bool vis[1010];
int head[3010],n,f,d,cnt=0;
void ins(int s,int t,int c) {
    e[++cnt].to = t;
    e[cnt].next = head[s];
    head[s] = cnt;
    e[cnt].cap = c;
    e[cnt].rev = cnt + 1;
    e[++cnt].to = s;
    e[cnt].next = head[t];
    head[t] = cnt;
    e[cnt].cap = 0;
    e[cnt].rev = cnt - 1;
}
int dfs(int v,int t,int f) {
    if (v == t)return f;
    vis[v] = 1;
    for (int i = head[v]; i; i = e[i].next) {
        if (!vis[e[i].to] && e[i].cap > 0) {
            int d = dfs(e[i].to, t, min(f, e[i].cap));
            if (d > 0) {
                e[i].cap -= d;
                e[e[i].rev].cap += d;
                return d;
            }
        }
    }
    return 0;
}
int max_flow(int s,int t) {
    int flow = 0;
    for (;;) {
        memset(vis, 0, sizeof(vis));
        int f = dfs(s, t, 1e9);
        if (f == 0)return flow;
        flow += f;
    }
}
int main() {
    while (~scanf("%d%d%d", &n, &f, &d)) {
        memset(e, 0, sizeof(e));
        memset(head, 0, sizeof(head));
        cnt = 0;
        int s = f + d + 1 + 2 * n, t = f + d + 2 + 2 * n;
        int x, y;
        for (int i = 1; i <= f; ++i)
            ins(s, i, 1);
        for (int i = 1; i <= d; ++i)
            ins(f + 2 * n + i, t, 1);
        for (int i = 1; i <= n; ++i)
            ins(f + i, f + i + n, 1);
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &x, &y);
            int num;
            for (int j = 1; j <= x; ++j) {
                scanf("%d", &num);
                ins(num, i + f, 1);
            }
            for (int j = 1; j <= y; ++j) {
                scanf("%d", &num);
                ins(f + n + i, f + n * 2 + num, 1);
            }
        }
        printf("%d\n", max_flow(s, t));
    }
    return 0;
}