UVa 11045 My T-shirt suits me

  My T-shirt suits me 

Our friend Victor participates as an instructor in an environmental volunteer program. His boss asked Victor to distribute N T-shirts to M volunteers, one T-shirt each volunteer, where N is multiple of six, and NM. There are the same number of T-shirts of each one of the six available sizes: XXL, XL, L, M , S, and XS. Victor has a little problem because only two sizes of the T-shirts suit each volunteer.

You must write a program to decide if Victor can distribute T-shirts in such a way that all volunteers get a T-shirt that suit them. If N  M, there can be some remaining T-shirts.

Input 

The first line of the input contains the number of test cases. For each test case, there is a line with two numbers N and M. N is multiple of 6, 1N36, and indicates the number of T-shirts. Number M, 1M30, indicates the number of volunteers, with NM. Subsequently, M lines are listed where each line contains, separated by one space, the two sizes that suit each volunteer (XXL, XL, L, M , S, or XS).

Output 

For each test case you are to print a line containing YES if there is, at least, one distribution where T-shirts suit all volunteers, or NO, in other case.

Sample Input 

 
3
18 6
L XL
XL L
XXL XL
S XS
M S
M L
6 4
S XL
L S
L XL
L XL
6 1
L M

Sample Output 

 
YES
NO
YES

 二分图匹配问题，将志愿者和就服匹配起来

直接套用匈牙利算法模板即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<vector>
#define MAX_V 80

using namespace std;

int V;
int n,m;
int match[MAX_V];
bool used[MAX_V];
vector<int> G[MAX_V];

void add_edge(int u,int v)
{
    G[u].push_back(v);
    G[v].push_back(u);
}

bool dfs(int v)
{
    used[v]=true;

    for(int i=0;i<G[v].size();i++)
    {
        int u=G[v][i],w=match[u];
        if(w<0||(!used[w]&&dfs(w)))
        {
            match[v]=u;
            match[u]=v;
            return true;
        }
    }

    return false;
}

int bipartite_matching()
{
    int res=0;

    memset(match,-1,sizeof(match));

    for(int v=1;v<=V;v++)
    {
        if(match[v]<0)
        {
            memset(used,false,sizeof(used));
            if(dfs(v))
                res++;
        }
    }

    return res;
}

int main()
{
    int kase;

    scanf("%d",&kase);

    while(kase--)
    {
        scanf("%d %d",&n,&m);

        V=n+m;
        for(int i=1;i<=V;i++)
            G[i].clear();

        string s1,s2;
        for(int i=1;i<=m;i++)
        {
            cin>>s1>>s2;
            if(s1=="XXL"||s2=="XXL")
                for(int j=1;j<=n/6;j++)
                    add_edge(j,n+i);
            if(s1=="XL"||s2=="XL")
                for(int j=n/6+1;j<=n/3;j++)
                    add_edge(j,n+i);
            if(s1=="L"||s2=="L")
                for(int j=n/3+1;j<=n/2;j++)
                    add_edge(j,n+i);
            if(s1=="M"||s2=="M")
                for(int j=n/2+1;j<=n*2/3;j++)
                    add_edge(j,n+i);
            if(s1=="S"||s2=="S")
                for(int j=n*2/3+1;j<=n*5/6;j++)
                    add_edge(j,n+i);
            if(s1=="XS"||s2=="XS")
                for(int j=n*5/6+1;j<=n;j++)
                    add_edge(j,n+i);
        }

        if(bipartite_matching()==m)
            puts("YES");
        else
            puts("NO");
    }

    return 0;
}