hihoCoder 1227 The Cats' Feeding Spots(暴力)——ACM-ICPC国际大学生程序设计竞赛北京赛区(2015)网络赛

题目1 : The Cats' Feeding Spots

时间限制: 1000ms

单点时限: 1000ms

内存限制: 256MB

描述

In Yan Yuan, the Peking University campus, there are many homeless cats. They all live happy lives because students founded a Cat Association to take care of them. Students not only feed them, but also treat their illness and sterilize some of them. Students make many feeding spots for the cats and cats always gather around those spots and make a lot of noise at night. Now the university authorities decide to restrict the number of feeding spots. This is the conversation between an officer and Rose Li, the director of Cat Association, and also a ACMer.

"Rose, From now on, you can't build any new feeding spots any more. But I want you to keep exactly N feeding spots, and you should make the area which contains the feeding spots as small as possible!"

"Oh, do you mean that we should find a smallest convex hull which contains N spots?"

"Convex hull? What is a convex hull? Please speak Chinese!"

"All right, forget the convex hull. So what do you mean the 'area', what's its shape?"

"It means... and the shape? Oh... let's do it this way: you can choose any feeding spot as center, and then draw a circle which includes exactly N spots. You should  find the smallest circle of such kind, and then we remove all feeding spots outside that circle."

Although this way sounds a little bit ridiculous, Rose still writes a program to solve the problem. Can you write the program?

输入

The first line is an integer T (T <= 50), meaning the number of test cases.

Then T lines follow, each describing a test case.

For each test case:

Two integer M and N go first(1 <= M, N <= 100), meaning that there are M feeding spots originally and Rose Li has to keep exactly N spots.

Then M pairs of real numbers follow, each means a coordinate of a feeding spot in Yan Yuan. The range of coordinates is between [-1000,1000]

输出

For each test case, print the radius of the smallest circle. Please note that the radius must be an POSITIVE INTEGER and no feeding spots should be located just on the circle because it's hard for the campus gardeners to judge whether they are inside or outside the circle.  If there are no solution, print "-1" instead.

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

/*********************************************************************/

题意：此题简单来说就是给你m个点的坐标和一个整数n，以某个点为圆心作最小圆，使得恰好包含n个点(不允许有点在圆上，而圆心的那个点也算包含的点)，若存在，输出此圆的半径，否则输出"-1"

解题思路：此题暴力就可以过了，因为半径的值是整数，所以枚举可能的半径值(因为坐标的取值范围是[-1000,1000]，故半径可能值最大为)以及作为圆心的点，并判断包含在圆内的点的个数是否恰好为n即可

另外此题数据是包含n>m的情况的

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<deque>
#include<cmath>
#include<functional>
#include<iterator>
#include<set>
#include<utility>
#include<stack>
#include<queue>
#include<iostream>
using namespace std;
#define maxn 110000
#define mod 530600414
double a[1005];
double b[1005];
int main()
{
    int m,i,j,k;
    scanf("%d",&m);
    while(m--)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        for(i=0; i<x; i++)
            scanf("%lf%lf",&a[i],&b[i]);
        int minn=1000000;
        int ans;
        for(i=1; i<=2900; i++)
        {
            for(j=0; j<x; j++)
            {
                ans=1;
                for(k=0; k<x; k++)
                {
                    if(k==j)
                        continue;
                    if(i*i>((a[j]-a[k])*(a[j]-a[k])+(b[j]-b[k])*(b[j]-b[k])))
                        ans++;
                    else if(i*i == ((a[j]-a[k])*(a[j]-a[k])+(b[j]-b[k])*(b[j]-b[k])))
                    {
                        ans=-1;
                        break;
                    }
                }
                if(ans==y)
                {
                    minn=i;
                    break;
                }
            }
            if(minn!=1000000)
                break;
        }
        if(minn!=1000000)
            printf("%d\n",minn);
        else
            printf("-1\n");
    }
    return 0;
}

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