Uva11045

题意:有6种类型的志愿服,给你n件衣服(n一定是6的倍数),然后m个志愿者,接下里有m行,有两个字符串,表示志愿者能穿的衣服的尺码,问你是否所有志愿者都能找到合适的衣服= =

最大流:0为源点,m+7为汇点,源点到每种志愿服的容量是n/6,志愿服和志愿者之间的容量是1,志愿者和汇点之间的容量是1,然后其余源点到汇点的最大流F,若F==m则输出YES,否则就是NO..

#include<stdio.h>
#include<string.h> 
#include<algorithm>
#include<iostream>
#include<string> 
#include<vector>
#include<queue>
const int inf=89999999;
using namespace std;
const int N=100;
string g[10]={"\0","XXL","XL","L","M","S","XS"};
int G[100][100];
int a[100];
int flow[100][100];
int p[100];
int n,m,F,V;
void EK_()
{
    queue<int>q;
    memset(flow,0,sizeof(flow));
    for(;;)
    {
        memset(a,0,sizeof(a)); 
        a[0]=inf; 
        q.push(0);
        while(!q.empty())
        {
            int u=q.front();
            q.pop();
            for(int v=0;v<=V;v++)
            { 
            if(!a[v]&&G[u][v]>flow[u][v])
            {
            p[v]=u;
            q.push(v);
            a[v]=a[u];
            if(a[v]>G[u][v]-flow[u][v])
            a[v]=G[u][v]-flow[u][v];
            }
        } 
        }    
        if(a[V]==0)
        break;
        for(int u=V;u!=0;u=p[u])
        {
            flow[p[u]][u]+=a[V];
            flow[u][p[u]]-=a[V];
        }
        F+=a[V];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
    memset(G,0,sizeof(G));
    scanf("%d%d",&n,&m);
    V=m+7;
    string s1,s2;
    for(int i=1;i<=6;i++)
    G[0][i]=n/6;
    for(int i=1;i<=m;i++)
    {
        cin>>s1>>s2;
        int a1,a2;
        for(a1=1;g[a1]!=s1;a1++);
        for(a2=1;g[a2]!=s2;a2++);
        G[a1][i+6]=1;
        G[a2][i+6]=1;
        G[i+6][V]=1;
    }
    F=0;
    EK_();
    if(F==m)
    cout<<"YES\n";
    else
    cout<<"NO\n";
}
return 0;
}

 

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