| Time Limit: 2000MS | Memory Limit: 32768K | |
| Total Submissions: 22296 | Accepted: 7040 |
Description
"How am I ever going to solve this PRoblem?" said the pilot.Indeed, the pilot was not facing an easy task. She had to drop packages at specific points scattered in a dangerous area. Furthermore, the pilot could only fly over the area once in a straight line, and she had to fly over as many points as possible. All points were given by means of integer coordinates in a two-dimensional space. The pilot wanted to know the largest number of points from the given set that all lie on one line. Can you write a program that calculates this number?Your program has to be efficient!Input
Input consist several case,First line of the each case is an integer N ( 1 < N < 700 ),then follow N pairs of integers. Each pair of integers is separated by one blank and ended by a new-line character. The input ended by N=0.Output
output one integer for each input case ,representing the largest number of points that all lie on one line.Sample Input
51 12 23 39 1010 110Sample Output
3題意：給你一個(gè)數(shù)n，代表在二維空間內(nèi)的n的點(diǎn)，找到在一條直線上最多的點(diǎn)能有多少。在空間內(nèi)的三個(gè)點(diǎn)共線，滿足（y1-y2）/ (x1-x2）==（y3-y4）/ （x3-x4） ，為了防止分母為0，化成（y1-y2）*（x3-x4）== (x1-x2）*（y3-y4）。用三重循環(huán)即可。#include<cstdio>#include <algorithm> #include <cstring> #define LL long longusing namespace std;struct Node{ int x,y;}a[705];int main(){ int n; while(scanf("%d",&n)&&n) { for(int i=0;i<n;i++) scanf("%d%d",&a[i].x,&a[i].y); int sum=0; for(int i=0;i<n-2;i++) { for(int j=i+1;j<n-1;j++) { int ans=2; for(int k=j+1;k<n;k++) { if((a[i].x-a[j].x)*(a[k].y-a[j].y)==(a[i].y-a[j].y)*(a[k].x-a[j].x)) ans++; } if(ans>sum) sum=ans; } } printf("%d/n",sum); } return 0;}
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