标签:blog os io for 问题 div amp size
把行和列都看做是点,小行星看成是边的话,那么这个显然就是求一个最小点覆盖集的问题。
最小点覆盖 == 最大匹配
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <climits> #include <string> #include <iostream> #include <map> #include <cstdlib> #include <list> #include <set> #include <queue> #include <stack> using namespace std; typedef long long LL; const int maxn = 505; int n,m,bx[maxn],by[maxn]; bool vis[maxn],g[maxn][maxn]; int dfs(int now) { for(int i = 1;i <= n;i++) if(!vis[i] && g[now][i]) { vis[i] = true; if(!by[i] || dfs(by[i])) { bx[now] = i; by[i] = now; return 1; } } return 0; } int main() { while(scanf("%d%d",&n,&m) != EOF) { memset(g,0,sizeof(g)); memset(bx,0,sizeof(bx)); memset(by,0,sizeof(by)); for(int i = 1;i <= m;i++) { int a,b; scanf("%d%d",&a,&b); g[a][b] = true; } int ans = 0; for(int i = 1;i <= n;i++) if(!bx[i]) { memset(vis,0,sizeof(vis)); if(dfs(i)) ans++; } printf("%d\n",ans); } return 0; }
POJ 3401 Asteroids 求最小点覆盖集,布布扣,bubuko.com
标签:blog os io for 问题 div amp size
原文地址:http://www.cnblogs.com/rolight/p/3881400.html