UVa 11045 - My T-shirt suits me(最大流)

题目转的有点远,不过如果对最大流问题比较熟悉的话也不难看出问题的本质。

下面描述建图过程。

1、设一个原点0,1到6为六种型号衣服对应的点,然后7到m+6是m个人对应的点,最后一个点m+7对应终结点。

2、原点到型号点的容量为n/6(根据题意),一个人对应两种型号衣服,每条线的容量都是1(每个人对应两种衣服),最后每个人到终点的容量都为1(每个人只需要一件衣服)。这样图就建好了。

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

const int N = 40;
const int inf = 1000000;
int n, m, T, s, t;
int cap[N][N], flow[N][N], p[N];

int maxFlow()
{
    queue<int> q;
    int ans = 0, a[N];
    while(true)
    {
        q.push(s);
        memset(a, 0, sizeof(a) );
        a[s] = inf;
        while (!q.empty())
        {
            int u = q.front(); q.pop();
            for ( int i = 0; i <= t; ++i )
            {
                if ( !a[i] && cap[u][i] > flow[u][i] )
                {
                    p[i] = u; q.push(i);
                    a[i] = min( a[u], cap[u][i] - flow[u][i] );
                }
            }
        }
        if ( a[t] == 0 )
            break;
        for ( int i = t; i != s; i = p[i] )
        {
            flow[p[i]][i] += a[t];
            flow[i][p[i]] -= a[t];
        }
        ans += a[t];
    }
    return ans;
}
int num( string s )
{
    if ( s == "XS" ) return 1;
    else if ( s == "S") return 2;
    else if ( s == "M") return 3;
    else if ( s == "L") return 4;
    else if ( s == "XL") return 5;
    else if ( s == "XXL") return 6;
}
int main()
{
    cin>>T;
    while ( T-- )
    {
        cin>>n>>m;
        s = 0, t = m + 7;
        int f = n / 6;
        memset(flow, 0, sizeof(flow));
        memset(cap, 0, sizeof(cap));
        for ( int i = 7; i <= 6 + m; ++i )
        {
            string a, b;
            cin >> a >> b;
            cap[num(a)][i] = 1;
            cap[num(b)][i] = 1;
            cap[i][t] = 1;
        }
        for ( int i = 1; i <= 6; ++i )
            cap[0][i] += f;
        int ans = maxFlow();
        if (ans < m)
            cout<<"NO"<<endl;
        else
            cout<<"YES"<<endl;
    }
    return 0;
}




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