hihocoder #1227 : The Cats' Feeding Spots(枚举)

#1227 : 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

43 2 0 0 1 0 1.2 02 2 0 0 1 02 1 0 0 1.2 02 1 0 0 1 0

样例输出

121-1

参考了别人的做法，原来不用都枚举的。先枚举点作为圆心，然后将圆心到其他点的距离求出来，这样免得将半径枚举1~1000了，那样真的好麻烦啊~。然后将前n个点纳入，然后处理半径为合适的整数，最后判定有木有纳多了。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;const int N = 1005;pair<double,double>p[N];double d[N];int main(){    int t,m,n;    cin >> t;    while(t--)    {        int ans = N;        cin >> m >> n;        for(int i = 0;i < m;i++)            scanf("%lf%lf",&p[i].first,&p[i].second);        for(int i = 0;i < m;i++)        {            for(int j = 0;j < m;j++)                d[j] = sqrt((p[i].first - p[j].first)*(p[i].first - p[j].first)+(p[i].second - p[j].second)*(p[i].second - p[j].second));            sort(d,d+m);            int R = ceil(d[n-1]);//覆盖0~n-1共n个点            R = R == d[n-1] ? R+1 : R;            if(m != n && R >= d[n])//第n+1个点压界，不符合条件m！=n否越界                continue;            if(R < ans)                ans = R;        }        if(ans == N)            printf("-1\n");        else            printf("%d\n",ans);    }    return 0;}