poj 2536 二分匹配

簡單二分匹配，注意，輸出的是n-最大匹配數

#include<stdio.h>#include<string.h>#include<math.h>int match[110];bool map[110][110];struct point{double x,y;}p[110],hole[110];double dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}bool vis[110];int m;bool dfs(int u){int i;for(i=0;i<m;i++){if(map[u][i]&&!vis[i]){vis[i]=true;if(match[i]==-1||dfs(match[i])){match[i]=u;return true;}}}return false;}int main(){int n,s,v;int i,j,k;while(scanf("%d%d%d%d",&n,&m,&s,&v)!=EOF){memset(map,false,sizeof(map));memset(match,-1,sizeof(match));for(i=0;i<n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);for(i=0;i<m;i++)scanf("%lf%lf",&hole[i].x,&hole[i].y);for(i=0;i<n;i++){for(j=0;j<m;j++){if(dis(p[i],hole[j])/v<=1.0*s){map[i][j]=true;}}}int ans=0;for(i=0;i<n;i++){memset(vis,false,sizeof(vis));if(dfs(i))ans++;}printf("%d\n",n-ans);}return 0;}