hdu 4619 Warm up 2_最大独立集

三个人整个下午都想不出这题

后来看题解，竟然用匈牙利算法的最大独立集，我顿时晕了。

题意:给竖着和横着的方块，除去重叠的，最多能留下几个方块

#include <cstdlib>
#include <iostream>
using namespace std;
#define N 1010
struct Point{
    int x,y;
    Point(){}
    Point(int a,int b){x=a;y=b;}
    bool operator==(const Point &a)const{
        return x==a.x&&y==a.y;
    }
}pn[N],pm[N];
int link[N],n,m;
bool vis[N],g[N][N];
bool judge(Point a,Point b){
    if(a==b||Point(a.x+1,a.y)==b||Point(b.x,b.y+1)==a||Point(a.x+1,a.y)==Point(b.x,b.y+1))
        return 1;
    return 0;
}
bool dfs(int u){
    int i;
    for(i=0;i<m;i++){
        if(!vis[i]&&g[u][i]){
            vis[i]=1;
            if(link[i]==-1||dfs(link[i])){
                link[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main(int argc, char *argv[])
{
    int i,j;
    while(scanf("%d%d",&n,&m)&&m||n){
        memset(g,0,sizeof(g));
        for(i=0;i<n;i++)
            scanf("%d%d",&pn[i].x,&pn[i].y);
        for(i=0;i<m;i++)
            scanf("%d%d",&pm[i].x,&pm[i].y);
        for(i=0;i<n;i++)
            for(j=0;j<m;j++)
                if(judge(pn[i],pm[j]))
                    g[i][j]=1;
        memset(link,-1,sizeof(link));
        int sum=0;
        for(i=0;i<n;i++){
            memset(vis,0,sizeof(vis));
            sum+=dfs(i);
        }
        printf("%d\n",n+m-sum);
    }
    //system("PAUSE");
    return EXIT_SUCCESS;
}