二分图最大匹配--poj2536

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Gopher II

Time Limit: 2000MS

Memory Limit: 65536K

Total Submissions: 5822

Accepted: 2428

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

题意：有很多洞和地鼠，如果地鼠在s秒内不能到洞里就会被吃掉，问被吃掉的最少的数目。
思路：二分图最大匹配，以地鼠和洞为节点建图，得出最多可以逃出的数目。

#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
struct node
{
    double x,y;
} gopher[105],hole[105];
int con[105][105];
bool vis[105];
int pre[105];
int n,m,s,v;
bool dist(node a,node b)
{
    double p=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
    if(p<=s*v)
        return 1;
    return 0;
}
bool dfs(int u)
{
    for(int i=1;i<=m;i++)
    {
        if((!vis[i])&&con[u][i])
        {
            vis[i]=1;
            if(pre[i]==0||dfs(pre[i]))
            {
                pre[i]=u;
                return 1;
            }
        }
    }
    return 0;
}
int main()
{
    //freopen("in.txt","r",stdin);
    while(cin>>n>>m>>s>>v)
    {
        memset(con,0,sizeof(con));
        memset(pre,0,sizeof(pre));
        for(int i=1; i<=n; i++)
            cin>>gopher[i].x>>gopher[i].y;
        for(int i=1; i<=m; i++)
            cin>>hole[i].x>>hole[i].y;
        int dis=s*v;
        for(int i=1; i<=n; i++)
            for(int j=1; j<=m; j++)
            {
                if(dist(gopher[i],hole[j]))
                    con[i][j]=1;
            }
        int ans=0;
        for(int i=1;i<=n;i++)
        {
            memset(vis,0,sizeof(vis));
            if(dfs(i))
                ++ans;
        }
        cout<<n-ans<<endl;
    }
    return 0;
}