Gopher II
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 6345 | Accepted: 2599 |
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
最大独立集：
1 //200K 47MS C++ 1200B 2014-06-10 07:43:56
2 #include<stdio.h>
3 #include<string.h>
4 #include<math.h>
5 #define N 105
6 struct node{
7 double x,y;
8 }gopher[N],hole[N];
9 int g[N][N];
10 int match[N];
11 int vis[N];
12 int n,m;
13 double dist(node a,node b)
14 {
15 return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
16 }
17 int dfs(int u)
18 {
19 for(int i=0;i<m;i++)
20 if(!vis[i] && g[u][i]){
21 vis[i]=1;
22 if(match[i]==-1 || dfs(match[i])){
23 match[i]=u;
24 return 1;
25 }
26 }
27 return 0;
28 }
29 int hungary()
30 {
31 int ret=0;
32 memset(match,-1,sizeof(match));
33 for(int i=0;i<n;i++){
34 memset(vis,0,sizeof(vis));
35 ret+=dfs(i);
36 }
37 return ret;
38 }
39 int main(void)
40 {
41 double s,v;
42 while(scanf("%d%d%lf%lf",&n,&m,&s,&v)!=EOF)
43 {
44 for(int i=0;i<n;i++)
45 scanf("%lf%lf",&gopher[i].x,&gopher[i].y);
46 for(int i=0;i<m;i++)
47 scanf("%lf%lf",&hole[i].x,&hole[i].y);
48 memset(g,0,sizeof(g));
49 for(int i=0;i<n;i++)
50 for(int j=0;j<m;j++)
51 if(dist(gopher[i],hole[j])/v<s)
52 g[i][j]=1;
53 printf("%d\n",n-hungary());
54 }
55 return 0;
56 }