正题

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大意

有n个地鼠有m个地洞，每只地鼠的速度都是v。若不在s秒之内回到地洞就会狗带，每只地洞只能躲一只地鼠，求能活下来多少只

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代码

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
struct line{int x,y,next;
}a[10001];
int link[101],n,m,ls[101],xx,yy,ss,t,w,si,vi;
double dx[101],dy[101],zx,zy;
double s,v;
bool cover[101];
double pows(double x)
{return x*x;
}
double dis(double x1,double y1,double x2,double y2)
{return sqrt(pows(x1-x2)+pows(y1-y2));
}//计算距离
bool find(int x)//寻找增广链
{int p=0;for (int q=ls[x];q;q=a[q].next){if (!cover[a[q].y]){p=link[a[q].y];link[a[q].y]=x;cover[a[q].y]=true;if (!p || find(p)) return true;link[a[q].y]=p;}}return false;
}
int main()
{scanf("%d",&t);for (int i=1;i<=t;i++){memset(link,0,sizeof(link));memset(ls,0,sizeof(ls));w=0;ss=0;scanf("%d%d%d%d",&n,&m,&si,&vi);s=si;v=vi;for (int i=1;i<=n;i++){scanf("%lf%lf",&dx[i],&dy[i]);}double q=0;for (int i=1;i<=m;i++){scanf("%lf%lf",&zx,&zy);for (int j=1;j<=n;j++){q=dis(dx[j],dy[j],zx,zy);if (q<=s*v)//可以跑到{a[++w].x=j;a[w].y=i;a[w].next=ls[j];ls[j]=w;}}}for (int i=1;i<=n;i++){memset(cover,false,sizeof(cover));if (!find(i)) ss++;}eprintf("%d\n",ss);}
}