【POJ 3041】Asteroids (最小点覆盖)

每次选择清除一行或者一列上的小行星。最少选择几次。

将行和列抽象成点，第i行为节点i+n，第j列为节点j，每个行星则是一条边，连接了所在的行列。

于是问题转化成最小点覆盖。二分图的最小点覆盖==最大匹配。

#include <cstdio>
#include <cstring>
const int N=505<<1;
int n,vis[N],link[N],g[N][N];
int find(int u)
{
	for(int i=1; i<=n*2; i++)
		if(g[u][i]&&!vis[i])
		{
			vis[i]=1;
			if(!link[i]||find(link[i]))
			{
				link[i]=u;
				return 1;
			}
		}
	return 0;
}
int solve(){
	memset(link,0,sizeof link);
	int ans=0;
	for(int i=1;i<=n;i++){
		memset(vis,0,sizeof vis);
		if(find(i))ans++;
	}
	return ans;
}
int main(){
	int m,u,v;
	scanf("%d%d",&n,&m);
	memset(g,0,sizeof g);
	while(m--){
		scanf("%d%d",&u,&v);
		g[u+n][v]=g[v][u+n]=1;
	}
	printf("%d\n",solve());
}

　　

时间： 2024-10-09 23:35:01

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