POJ 3281 Dining (最大流)

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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:巧在建图

见白书：

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#include<vector>using namespace std;const int maxn = 505;const int INF = 0x3f3f3f3f;int g[maxn][maxn], n, f, d;int path[maxn], flow[maxn];int start, endd;//f+1 - f+1+n 食物一侧的牛//f+n+1 - f+2n+1 饮料一侧的牛//1 - f 食物//f+2n+1 - f+2n+d 饮料//start:0, enddd:f+2n+d+1int bfs(){    queue<int> q;    memset(path, -1, sizeof(path));    path[start] = 0, flow[start] = INF;    q.push(start);    while(!q.empty())    {        int t = q.front(); q.pop();        if(t == endd) break;        for(int i = 0; i <= endd; i++)        {            if(path[i] == -1 && g[t][i])            {                flow[i] = min(flow[t], g[t][i]);                q.push(i);                path[i] = t;            }        }    }    if(path[endd] == -1) return -1;    return flow[endd];}int E_K(){    int max_flow = 0, step, now, pre;    while((step=bfs()) != -1)    {        max_flow += step;        now = endd;        while(now != start)        {            pre = path[now];            g[pre][now] -= step;            g[now][pre] += step;            now = pre;        }    }    return max_flow;}int main(void){    while(cin >> n >> f >> d)    {        start = 0;        endd = f+d+2*n+1;        memset(g, 0, sizeof(g));        //start - food        for(int i = 1; i <= f; i++)            g[0][i] = 1;        //cow1 - cow2        for(int i = f+1; i <= f+n; i++)            g[i][i+n] = 1;        //drink - endd        for(int i = f+2*n+1; i <= f+2*n+d; i++)            g[i][endd] = 1;        for(int i = 0; i < n; i++)        {            int x, numf, numd;            scanf("%d%d", &numf, &numd);            for(int j = 0; j < numf; j++)            {                scanf("%d", &x);                g[x][f+1+i] = 1;            }            for(int j = 0; j < numd; j++)            {                scanf("%d", &x);                g[f+n+1+i][f+2*n+x] = 1;            }        }        printf("%d\n", E_K());    }    return 0;}

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