POJ 2536 最大匹配入门

Gopher II

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3680		Accepted: 1539

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

Source
Waterloo local 2001.01.27
敲错了一个变量，让我WA了很久都没看出来。。
Source Code

Problem: 2536		User: bingshen
Memory: 748K		Time: 32MS
Language: G++		Result: Accepted

Source Code

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<math.h>
#include<vector>

using namespace std;

struct Point
{
double x;
double y;
};

Point p1[105];
Point p2[105];
vector<int>map[105];
int link[105];
bool cover[105];
int n,m;

void init()
{
memset(link,-1,sizeof(link));
memset(map,0,sizeof(map));
}

double dis(Point a,Point b)
{
return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

void makemap(int s,int v)
{
int i,j;
double r;
r=s*v;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(dis(p1[i],p2[j])<=r)
map[i].push_back(j);
}
}

bool find(int v)
{
int i;
for(i=0;i<map[v].size();i++)
{
if(!cover[map[v][i]])
{
cover[map[v][i]]=true;
if(link[map[v][i]]==-1||find(link[map[v][i]]))
{
link[map[v][i]]=v;
return true;
}
}
}
return false;
}

int main()
{
int i,life;
double s,v;
while(scanf("%d%d%lf%lf",&n,&m,&s,&v)!=EOF)
{
init();
life=0;
for(i=0;i<n;i++)
scanf("%lf%lf",&p1[i].x,&p1[i].y);
for(i=0;i<m;i++)
scanf("%lf%lf",&p2[i].x,&p2[i].y);
makemap(s,v);
for(i=0;i<n;i++)
{
memset(cover,0,sizeof(cover));
if(find(i))
life++;
}
printf("%d/n",n-life);
}
return 0;
}