poj 3281 Dining(网络流最大流)

题意：有F种食物和D种饮料，每种食物或饮料只能供一头牛享用，且每头牛只享用一种食物和一种饮料。现在有n头牛，每头牛都有自己喜欢的食物种类列表和饮料种类列表，问最多能使几头牛同时享用到自己喜欢的食物和饮料。（1 <= f <= 100, 1 <= d <= 100, 1 <= n <= 100）
思路：EK算法，拆点。

参考博客：http://blog.sina.com.cn/s/blog_6635898a0100ly5j.html
                  http://blog.csdn.net/zhang20072844/article/details/8122626  

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#define inf 1000
#define nMax 410
#define Max(a,b) (a>b?a:b)
#define Min(a,b) (a<b?a:b)

int map[nMax][nMax];
int N,F,D;
int path[nMax];
int queue[nMax * 100];
int head,end;
//bool flag[nMax];
//广搜求一条增广路
int bfs()
{
    int minFlow = inf,u;
    memset(path, -1, sizeof(path));
    head = 0;
    end = 1;
    queue[head] = 0;
    while (head < end)
    {
        u = queue[head ++];
        if (u == 2 * N + F + D + 1)
        {
            break;
        }
        for (int i = 1; i <= 2 * N + F + D + 1; ++ i)
        {
            if (path[i] == -1 && map[u][i] )
            {
                if (minFlow > map[u][i])
                {
                    minFlow = map[u][i];
                }
                queue[end ++] = i;
                path[i] = u;

            }
        }
    }
    if (path[2 * N + F + D + 1] == -1)
    {
        return -1;
    }
    return minFlow;
}
//EK算法，每次广搜得到一条增广路径，然后更新残留网络
void Edmods_Karp()
{
    int flow, maxFlow = 0, now, pre;
    while ((flow = bfs()) != -1)
    {
        maxFlow += flow;
        now = 2 * N + F + D + 1;
        while (now != 0)
        {
            pre = path[now];
            map[pre][now] -= flow;
            map[now][pre] += flow;
            now = pre;
        }
    }
    printf("%d\n", maxFlow);
}
//按照源点-食物-牛-牛-饮料-汇点的顺序建图
void buildMap()
{
    int fNum,dNum,fd;
    while (scanf("%d %d %d", &N, &F, &D) != EOF)
    {
        memset(map, 0, sizeof(map));
        //memset(flag, false, sizeof(flag));
        for (int i = 1; i <= N; ++ i)
        {
            scanf("%d %d", &fNum, &dNum);
            for (int j = 0; j < fNum; ++ j)
            {
                scanf("%d", &fd);
                map[0][fd] = 1;
                map[fd][i + F] = 1;
            }
            map[i + F][i + F + N] = 1;
            for (int j = 0; j < dNum; ++ j)
            {
                scanf("%d", &fd);
                map[fd + 2 * N + F][F + 2 * N + D + 1] = 1;
                map[i + F + N][fd + 2 * N + F] = 1;
            }
        }
        Edmods_Karp();
    }
}
//注意这里给点编号，0-源点，1-F是食物，F+1-F+N是牛左点，F+N+1-F+N+N是牛右点，F+N+N+1-F+N+N+D是drink饮料点，F+N+N+D+1是汇点

int main()
{
    buildMap();
    return 0;
}