hiho 1227 找到一个恰好包含n个点的圆 （2015北京网赛 A题）

平面上有m个点，要从这m个点当中找出n个点，使得包含这n个点的圆的半径（圆心为n个点当中的某一点且半径为整数）最小，同
时保证圆周上没有点。

n > m 时要输出-1

样例输入
4
3 2 0 0 1 0 1.2 0
2 2 0 0 1 0
2 1 0 0 1.2 0
2 1 0 0 1 0
样例输出
1
2
1
-1

 

 1 # include <iostream>
 2 # include <cstdio>
 3 # include <cstring>
 4 # include <algorithm>
 5 # include <string>
 6 # include <cmath>
 7 # include <queue>
 8 # include <list>
 9 # define LL long long
10 using namespace std ;
11 
12 double dis[200][200] ;
13 const int INF = 0x3f3f3f3f ;
14 
15 struct Point
16 {
17     double x,y;
18 
19 }p[200];
20 
21 double dist(Point a,Point b)
22 {
23     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
24 }
25 
26 int main()
27 {
28     //freopen("in.txt","r",stdin) ;
29     int T ;
30     scanf("%d" , &T) ;
31     while(T--)
32     {
33         int m , n ;
34         int i , j ;
35         scanf("%d %d" , &m , &n) ;
36         for (i = 0 ; i < m ; i++)
37             scanf("%lf %lf" , &p[i].x , &p[i].y) ;
38         if (n > m)
39         {
40             printf("-1\n") ;
41             continue ;
42         }
43         int ans = INF ;
44         int r ;
45         memset(dis , 0 , sizeof(dis)) ;
46         for (i = 0 ; i < m ; i++)
47         {
48             for (j = i+1 ; j < m ; j++)
49             {
50                 dis[i][j] = dis[j][i] = dist(p[i] , p[j]) ;
51             }
52             sort(dis[i] , dis[i]+m) ;
53             r = (int)dis[i][n-1] ;
54             if (r <= dis[i][n-1])
55                 r += 1 ;
56             if (n < m &&r >= dis[i][n])
57                 continue ;
58             if (r < ans)
59                 ans = r ;
60         }
61         if (ans == INF)
62             printf("-1\n") ;
63         else
64             printf("%d\n" , ans) ;
65 
66     }
67     return 0;
68 }
View Code

 

posted @ 2015-10-01 22:51  __Meng  阅读(248)  评论(0)  编辑  收藏  举报