poj 2446 二分图最大匹配

思路：由(i+j)为偶数的点向(i+j)为奇数的点建边。求一次最大匹配，若正好为空格数(不包含洞)的一半，即输出YES。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define Maxn 1101
using namespace std;
int n,m,vi[Maxn],match[Maxn],graphic[Maxn][Maxn],map[40][40],N[Maxn],M[Maxn],x,y;
int dfs(int u)
{
    int i;
    for(i=1;i<=y;i++)
    {
        if(!vi[M[i]]&&graphic[u][M[i]])
        {
            vi[M[i]]=1;
            if(match[M[i]]==-1||dfs(match[M[i]]))
            {
                match[M[i]]=u;
                return 1;
            }
        }
    }
    return 0;
}
int Pos(int i,int j)
{
    return (i-1)*n+j;
}
int main()
{
    int i,j,t,a,b,k;
    while(scanf("%d%d%d",&m,&n,&k)!=EOF)
    {
        
        memset(match,-1,sizeof(match));
        memset(graphic,0,sizeof(graphic));
        memset(map,0,sizeof(map));
        for(i=1;i<=k;i++)
        {
            scanf("%d%d",&a,&b);
            map[b][a]=1;
        }
        x=y=0;
        for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
            {
                if((i+j)%2)
                    M[++y]=Pos(i,j);
                else
                    N[++x]=Pos(i,j);
                if(j<n)
                {
                    if(!map[i][j+1]&&!map[i][j])
                    {
                        if((i+j)%2==0)
                        graphic[Pos(i,j)][Pos(i,j+1)]=1;
                        if((i+j)%2)
                        {
                        graphic[Pos(i,j+1)][Pos(i,j)]=1;
                        }
                    }
                }
                if(i<m)
                {
                    if(!map[i+1][j]&&!map[i][j])
                    {
                        if((i+j)%2)
                        graphic[Pos(i+1,j)][Pos(i,j)]=1;
                        if((i+j)%2==0)
                        graphic[Pos(i,j)][Pos(i+1,j)]=1;
                    }
                }
            }
        }
        int ans=n*m-k;
        if(ans%2==1)
        {
            printf("NO\n");
            continue;
        }
        int num=0;
        for(i=1;i<=x;i++)
        {
            memset(vi,0,sizeof(vi));
            if(dfs(N[i]))
            {
                num++;
            }
        }
        if(num*2==ans)
            printf("YES\n");
        else
            printf("NO\n");

    }
    return 0;
}