SOJ 4021 Cocircular Points

题目连接：http://zuojie.3322.org:88/soj/problem.action?id=4021

题目：

Description

You probably know what a set of collinear points is: a set of points such that there exists astraight line that passes through all of them. A set of cocircular points is defined in the samefashion, but instead of a straight line, we ask that there is a circle such that every point of theset lies over its perimeter.The International Collinear Points Centre (ICPC) has assigned you the following task: given aset of points, calculate the size of the larger subset of cocircular points.

Input

Each test case is given using several lines. The first line contains an integer N representing thenumber of points in the set (1 <= N <= 100). Each of the next N lines contains two integers Xand Y representing the coordinates of a point of the set (-10^4 <= X, Y <= 10^4). Within eachtest case, no two points have the same location.The last test case is followed by a line containing one zero.

Output

For each test case output a single line with a single integer representing the number of pointsin one of the largest subsets of the input that are cocircular.

Sample Input

7-10 00 -1010 00 10-20 10-10 20-2 44-10000 1000010000 1000010000 -10000-10000 -99993-1 00 01 00

Sample Output

532SourceLatin American Regional Contest 2010-->

 

直接枚举三个点，然后去计算圆心的方法是会超时的，标准的方法是去枚举两个点，然后在枚举第三个点，看构成的圆的圆心有多少重复再了一个地方。

通过统计这些次数我们便可以知道最多有多少个点共圆。

时间复杂度也就可以从O(n^4)降到了O(n^3logn)

虽然还是很高，但是适当的剪枝（比如枚举的时候只需要枚举一半左右的点）

还是可以过的。

 

我的代码：

#include<stdio.h>#include<algorithm>#include<math.h>#define eps 1e-8using namespace std;struct node{double x;double y;};struct line{node a;node b;};node p[105];int n;bool cmp(node a,node b){if((a.x<b.x)||((fabs(a.x-b.x)<=eps)&&(a.y<b.y)))return true;return false;}double xmult(node p1,node p2,node p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}bool judge_line(int a,int b,int c){double temp;temp=xmult(p[a],p[b],p[c]);if(fabs(temp)<=eps)return false;elsereturn true;}node circle(node p1,node p2,node p3){    node p;    double a1=2*(p1.x-p2.x);    double a2=2*(p1.x-p3.x);    double b1=2*(p1.y-p2.y);    double b2=2*(p1.y-p3.y);    double c1=p1.x*p1.x+p1.y*p1.y-p2.x*p2.x-p2.y*p2.y;    double c2=p1.x*p1.x+p1.y*p1.y-p3.x*p3.x-p3.y*p3.y;    double t=a1*b2-a2*b1;p.x=(c1*b2-c2*b1)/t;p.y=(a1*c2-a2*c1)/t;    return p;}int getans(int a,int b){int i,num=0,kkk=3,ret=0;double dis;node center[105];for(i=b+1;i<=n;i++){if(judge_line(i,a,b))center[num++]=circle(p[a],p[b],p[i]);}if(num==0)return 2;sort(center,center+num,cmp);for(i=1;i<num;i++){dis=(center[i-1].x-center[i].x)*(center[i-1].x-center[i].x)+(center[i-1].y-center[i].y)*(center[i-1].y-center[i].y);if(dis<=eps)kkk++;else{if(kkk>ret)ret=kkk;kkk=3;}}if(kkk>ret)ret=kkk;return ret;}int main(){int i,j,temp,max;while(scanf("%d",&n)!=EOF){if(n==0)break;max=0;for(i=1;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);for(i=1;i<=n;i++)for(j=i+1;j<=n;j++){temp=getans(i,j);if(temp>max)max=temp;}printf("%d\n",max);}return 0;}