北京赛区(2015)网络赛The Cats' Feeding Spots

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

解题思路：枚举每一个点，二分查找半径（注意凡是有点在圆上的半径都不能要）

#include<iostream>#include<cstdio>#include<cmath>#include<queue>#include<algorithm>#include<stack>#include<cstring>#include<vector>#include<list>#include<set>#include<string>#include<map>using namespace std;int main(){    int t;    cin>>t;    while(t--)    {        int m,n;        scanf("%d%d",&m,&n);        double ma[105][5];        double dis[105][105];        for(int i=1; i<=m; i++)        {            scanf("%lf%lf",&ma[i][1],&ma[i][2]);        }        for(int i=1; i<=m; i++)        {            for(int j=1; j<=m; j++)            {                dis[i][j]=(ma[i][1]-ma[j][1])*(ma[i][1]-ma[j][1])+(ma[i][2]-ma[j][2])*(ma[i][2]-ma[j][2]);            }        }        int minr=10000;        for(int i=1; i<=m; i++)        {            int r=minr,l=1,s=0;            while(r-l>1)            {                int mid=(r+l)/2;                int x=0;                for(int j=1; j<=m; j++)                {                        if(dis[i][j]<mid*mid) x++;                }                if(x>=n) r=mid;                else l=mid;            }           // cout<<"r="<<r<<"      l="<<l<<endl;            int x=0;            bool mark=true;            for(int j=1; j<=m; j++)            {                    if(dis[i][j]<l*l) x++;                    if(dis[i][j]==l*l) {                            mark=false;                    break;                    }            }            if(x==n&&mark) minr=min(minr,l);            //cout<<"minr="<<minr<<endl;            x=0;            mark=true;            for(int j=1; j<=m; j++)            {                if(dis[i][j]<r*r) x++;                if(dis[i][j]==r*r){mark=false;break;}            }            if(x==n&&mark) minr=min(minr,r);            //cout<<"minr="<<minr<<endl;        }        if(minr==10000)           {               printf("-1\n");           }        else printf("%d\n",minr);    }    return  0;}