POJ 3281 Dining (最大流)

Posted 2020-07-30 季末Despaire

Dining

Time Limit: 2000MS		Memory Limit: 65536K

Description

Cows are such finicky eaters. Each cow has a preference for certain foods and drinks, and she will consume no others.

Farmer John has cooked fabulous meals for his cows, but he forgot to check his menu against their preferences. Although he might not be able to stuff everybody, he wants to give a complete meal of both food and drink to as many cows as possible.

Farmer John has cooked F (1 ≤ F ≤ 100) types of foods and prepared D (1 ≤ D ≤ 100) types of drinks. Each of his N (1 ≤ N ≤ 100) cows has decided whether she is willing to eat a particular food or drink a particular drink. Farmer John must assign a food type and a drink type to each cow to maximize the number of cows who get both.

Each dish or drink can only be consumed by one cow (i.e., once food type 2 is assigned to a cow, no other cow can be assigned food type 2).

Input

Line 1: Three space-separated integers: N, F, and D 
Lines 2..N+1: Each line i starts with a two integers F_i and D_i, the number of dishes that cow i likes and the number of drinks that cow i likes. The next F_i integers denote the dishes that cow i will eat, and the D_i integers following that denote the drinks that cow i will drink.

Output

Line 1: A single integer that is the maximum number of cows that can be fed both food and drink that conform to their wishes

Sample Input

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

Sample Output

3

分析
之前做过类似的题了，建图就是将奶牛拆成两个点，左边连食物，右边连饮料，食物和饮料分别连源点和汇点，跑最大流就行了。

#include<stdio.h>
#include<string.h>
#include<iostream>
#include<queue>
using namespace std;
//****************************************************
//最大流模板Edmonds_Karp算法
//初始化：G[][],st,ed
//******************************************************
const int MAXN = 500;
const int INF = 0x3fffffff;
int G[MAXN][MAXN];//存边的容量，没有边的初始化为0
int path[MAXN],flow[MAXN],st,ed;
int n;//点的个数，编号0~n，n包括了源点和汇点

queue<int>q;
int bfs()
{
    int i,t;
    while(!q.empty()) q.pop();//清空队列
    memset(path,-1,sizeof(path));//每次搜索前都把路径初始化成-1
    path[st]=0;
    flow[st]=INF;//源点可以有无穷的流流进
    q.push(st);
    while(!q.empty()){
        t=q.front();
        q.pop();
        if(t==ed) break;
        for(i=0;i<=n;i++){
            if(i!=st&&path[i]==-1&&G[t][i]){
                flow[i]=flow[t]<G[t][i]?flow[t]:G[t][i];
                q.push(i);
                path[i]=t;
            }
        }
    }
    if(path[ed]==-1) return -1;//即找不到汇点上去了。找不到增广路径了
    return flow[ed];
}

int Edmonds_Karp()
{
    int max_flow=0;
    int step,now,pre;
    while((step=bfs())!=-1){
        max_flow+=step;
        now=ed;
        while(now!=st){
            pre=path[now];
            G[pre][now]-=step;
            G[now][pre]+=step;
            now=pre;
        }
    }
    return max_flow;
}

int main()
{
    int N,F,D;
    memset(G,0,sizeof(G));
    scanf("%d%d%d",&N,&F,&D);
    n=2*N+F+D+1;
    st=0,ed=n;
    for(int i=1;i<=F;i++) G[st][i]=1;
    for(int i=1;i<=N;i++) G[F+i][N+F+i]=1;
    for(int i=1;i<=D;i++) G[F+2*N+i][ed]=1;
    for(int i=1;i<=N;i++){
        int k1,k2,u;
        scanf("%d%d",&k1,&k2);
        for(int j=1;j<=k1;j++){
            scanf("%d",&u);
            G[u][F+i]=1;
        }
        for(int j=1;j<=k2;j++){
            scanf("%d",&u);
            G[F+N+i][2*N+F+u]=1;
        }
    }
    printf("%d\n",Edmonds_Karp());

    return 0;
}