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;

}