Gopher II（二分图最大匹配）

1538:Gopher II

总时间限制: 

2000ms 

内存限制: 

65536kB

描述

The gopher family, having averted the canine threat, must face a new predator. 

The are n gophers and m gopher holes, each at distinct (x, y) coordinates. A hawk arrives and if a gopher does not reach a hole in s seconds it is vulnerable to being eaten. A hole can save at most one gopher. All the gophers run at the same velocity v. The gopher family needs an escape strategy that minimizes the number of vulnerable gophers.

输入

The input contains several cases. The first line of each case contains four positive integers less than 100: n, m, s, and v. The next n lines give the coordinates of the gophers; the following m lines give the coordinates of the gopher holes. All distances are in metres; all times are in seconds; all velocities are in metres per second.

输出

Output consists of a single line for each case, giving the number of vulnerable gophers.

样例输入

2 2 5 101.0 1.02.0 2.0100.0 100.020.0 20.0

样例输出

1

来源

Waterloo local 2001.01.27

我理解的二分图最大匹配：

把集合A中的点和集合B中的点最大匹配，用A中的每个点和B中的点连线，如果这个点没有连过，就连上好啦，要是连过了，就往前找，看看前面连过的点能不能换一个连线，给这个点腾个地方。。。（如此直白。。。）

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;struct dot{double x,y;};dot gopher[150],hole[150];int n,m,s,v,d;double dis(int a,int b){return (gopher[a].x-hole[b].x)*(gopher[a].x-hole[b].x)+(gopher[a].y-hole[b].y)*(gopher[a].y-hole[b].y);}bool vis[150];int match[150];bool dfs(int x){for(int i=1;i<=m;i++){if(!vis[i]&&dis(x,i)<=(double)d*d){vis[i]=1;if(match[i]==-1||dfs(match[i])){match[i]=x;return true;}}}return false;}int ans(){memset(match,-1,sizeof(match));int sum=0;for(int i=1;i<=n;i++){memset(vis,0,sizeof(vis));if(dfs(i)){sum++;}}return sum;}int main(){while(scanf("%d %d %d %d",&n,&m,&s,&v)!=EOF){d=s*v;for(int i=1;i<=n;i++){scanf("%lf %lf",&gopher[i].x,&gopher[i].y);}for(int i=1;i<=m;i++){scanf("%lf %lf",&hole[i].x,&hole[i].y);}printf("%d\n",n-ans());}return 0;}