POJ 1274 The Perfect Stall （网络流-最大流）

The Perfect Stall

Description

Farmer John completed his new barn just last week, complete with all the latest milking technology. Unfortunately, due to engineering problems, all the stalls in the new barn are different. For the first week, Farmer John randomly assigned cows to stalls, but
it quickly became clear that any given cow was only willing to produce milk in certain stalls. For the last week, Farmer John has been collecting data on which cows are willing to produce milk in which stalls. A stall may be only assigned to one cow, and,
of course, a cow may be only assigned to one stall.

Given the preferences of the cows, compute the maximum number of milk-producing assignments of cows to stalls that is possible.

Input

The input includes several cases. For each case, the first line contains two integers, N (0 <= N <= 200) and M (0 <= M <= 200). N is the number of cows that Farmer John has and M is the number of stalls in the new barn. Each of the following N lines corresponds
to a single cow. The first integer (Si) on the line is the number of stalls that the cow is willing to produce milk in (0 <= Si <= M). The subsequent Si integers on that line are the stalls in which that cow is willing to produce milk. The stall numbers will
be integers in the range (1..M), and no stall will be listed twice for a given cow.

Output

For each case, output a single line with a single integer, the maximum number of milk-producing stall assignments that can be made.

Sample Input

5 5
2 2 5
3 2 3 4
2 1 5
3 1 2 5
1 2

Sample Output

4

Source

USACO 40

题目大意：

有n个奶牛和m个仓库，现在每个奶牛有自己喜欢去的仓库，并且它们只会去自己喜欢的仓库吃东西，问最多有多少奶牛能够吃到东西？

解题思路：

比较裸的1道网络流题目。

解题代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <cmath>
#include <algorithm>
using namespace std;

const int INF=(1<<30);
const int maxn=510,maxm=100000;

struct edge{
    int u,v,f,next;
    edge(int u0=0,int v0=0,int f0=0){
        u=u0;v=v0;f=f0;
    }
}e[maxm];

int src,sink,cnt,head[maxn];

void adde(int u,int v,int f){
    e[cnt].u=u,e[cnt].v=v,e[cnt].f=f,e[cnt].next=head[u],head[u]=cnt++;
    e[cnt].u=v,e[cnt].v=u,e[cnt].f=0,e[cnt].next=head[v],head[v]=cnt++;
}

void init(){
    cnt=0;
    memset(head,-1,sizeof(head));
}

queue <int> q;
bool visited[maxn];
int dist[maxn];

void bfs(){
    memset(dist,0,sizeof(dist));
    while(!q.empty()) q.pop();
    visited[src]=true;
    q.push(src);
    while(!q.empty()){
        int s=q.front();
        q.pop();
        for(int i=head[s];i!=-1;i=e[i].next){
            int d=e[i].v;
            if(e[i].f>0 && !visited[d]){
                q.push(d);
                dist[d]=dist[s]+1;
                visited[d]=true;
            }
        }
    }
}

int dfs(int u,int delta){
    if(u==sink) return delta;
    else{
        int ret=0;
        for(int i=head[u];delta && i!=-1;i=e[i].next){
            if(e[i].f>0 && dist[e[i].v]==dist[u]+1){
                int d=dfs(e[i].v,min(e[i].f,delta));
                e[i].f-=d;
                e[i^1].f+=d;
                delta-=d;
                ret+=d;
            }
        }
        return ret;
    }
}

int maxflow(){
    int ret=0;
    while(true){
        memset(visited,false,sizeof(visited));
        bfs();
        if(!visited[sink]) return ret;
        ret+=dfs(src,INF);
    }
    return ret;
}

int n,m;

void input(){
    init();
    src=0;
    sink=n+m+1;
    int cnt,x;
    for(int i=1;i<=n;i++) adde(src,i,1);
    for(int i=1;i<=n;i++){
        scanf("%d",&cnt);
        while(cnt-- >0){
            scanf("%d",&x);
            adde(i,x+n,1);
        }
    }
    for(int i=1;i<=m;i++) adde(n+i,sink,1);
}

void solve(){
    printf("%d\n",maxflow());
}

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        input();
        solve();
    }
    return 0;
}

POJ 1274 The Perfect Stall （网络流-最大流）