poj 2536 Gopher II

Description

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.

Input

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.

Output

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

Sample Input

2 2 5 10
1.0 1.0
2.0 2.0
100.0 100.0
20.0 20.0

Sample Output

1

Key To Problem

题意：有许多的地鼠和许多的地鼠坑，每个地鼠坑只能装一只地鼠，给定每个地鼠和地鼠坑的坐标，求在一定时间内有几只地鼠回到地鼠坑。
题解：裸的二分图匹配，首先将每只地鼠与其在规定时间所能达到的地鼠坑连边，之后用匈牙利算法求解有多少只地鼠可以在规定时间内回到地鼠坑，用n减去即为所求。

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define N 210
using namespace std;
struct ss
{
    int next,to;
};
ss Edge[N*N];
struct node
{
    double x,y;
};
node st[N];
node en[N];
int n,m,tot,s,v;
int head[N];
int f[N];
bool used[N];

void clear()
{
    memset(head,0,sizeof(head));
    tot=0;
}

void add(int x,int y)
{
    Edge[++tot].next=head[x];
    Edge[tot].to=y;
    head[x]=tot;
}

bool cmp(double x1,double y1,double x2,double y2)
{
    double p=sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    if(p<=s*v)return true;
    else return false;
}

bool dfs(int u)
{
    for(int i=head[u];i;i=Edge[i].next)
    {
        int to=Edge[i].to;
        if(!used[to])
        {
            used[to]=true;
            if(f[to]==-1||dfs(f[to]))
            {
                f[to]=u;
                return true;
            }
        }
    }
    return false;
}

int hungary()
{
    int cnt=0;
    memset(f,-1,sizeof(f));
    for(int i=1;i<=n;i++)
    {
        memset(used,0,sizeof(used));
        if(dfs(i))cnt++;
    }
    return cnt;
}

int main()
{
    while(scanf("%d%d%d%d",&n,&m,&s,&v)!=EOF)
    {
        clear();
        for(int i=1;i<=n;i++)scanf("%lf%lf",&st[i].x,&st[i].y);
        for(int i=1;i<=m;i++)scanf("%lf%lf",&en[i].x,&en[i].y);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                if(cmp(st[i].x,st[i].y,en[j].x,en[j].y))add(i,j);
        printf("%d\n",n-hungary());
    }
    return 0;
}