poj2780Linearity【共线】

最新推荐文章于 2017-04-25 00:43:06 发布

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最新推荐文章于 2017-04-25 00:43:06 发布

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分类专栏：计算几何文章标签： poj2780Linearity共线

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Linearity

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 8242		Accepted: 1897

Description

Alice often examines star maps where stars are represented by points in a plane and there is a Cartesian coordinate for each star. Let the Linearity of a star map be the maximum number of stars in a straight line.

For example, look at the star map shown on the figure above, the Linearity of this map is 3, because the star 1, star 2, and star 5 are within the same straight line, and there is no straight line that passes 4 stars.

You are to write a program to find the Linearity of a star map.

Input

Input will contain multiple test cases. Each describes a star map.

For each test case, the first line of the input contains the number of stars N (2 <= N <= 1000). The following N lines describe coordinates of stars (two integers X and Y per line separated by a space, 0 <= X, Y <= 1000). There can be only one star at one point of the plane.

Output

Output the Linearity of the map in a single line.

Sample Input

Sample Output

题意：求一条直线与平面最多的点数相交；

本来还想用我写lightoj1292的代码来个1A却TLE了然后各种优化还是TLE无奈看了别人的思路固定一点求它到其他个点的斜率然后计算斜率相等的最大数目即可

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define inf 1e20
#define eps 1e-10
using namespace std;
struct point{
	double x,y;
}A[1010];
int seg(double n){
	if(fabs(n)<eps)return 0;
	else if(n<0)return -1;
	return 1;
}
int MAX(int a,int b){
	return a>b?a:b;
}
double K[1010];
bool cmp(double a,double b){
	return a<b;
}
double Slop(point a,point b){
	if(seg(a.x-b.x)==0)return inf;
	return (a.y-b.y)/(a.x-b.x);
}
int main()
{
	int n,i,j,k,ans,temp,m;
	while(scanf("%d",&n)!=EOF){
		for(i=0;i<n;++i){
			scanf("%lf%lf",&A[i].x,&A[i].y);
		}
		for(ans=0,m=0,i=0;i<n;++i,m=0){
			for(j=i+1;j<n;++j){
				K[m++]=Slop(A[i],A[j]);
			}
			sort(K,K+m,cmp);temp=1;
			for(int v=1;v<m;++v){
				if(seg(K[v]-K[v-1])==0)temp++;
				else{
					ans=MAX(ans,temp);
					temp=1;
				}
			}
			ans=MAX(ans,temp);
		}
		printf("%d\n",ans+1);
	}
	return 0;
}