Lining Up
| Time Limit: 2000MS | Memory Limit: 32768K | |
| Total Submissions: 16372 | Accepted: 5145 |
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!
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
5 1 1 2 2 3 3 9 10 10 11 0
Sample Output
3
#include<iostream>#include<algorithm>using namespace std;double k[1002];double x[1002],y[1002];int main(){ int n; int i,j,p; int cnt; int num,max_num; while(cin>>n) { for(i=0;i<n;i++) { cin>>x[i]; cin>>y[i]; } num=2; max_num=2; for(i=0;i<n;i++) { cnt=0; for(j=0;j<n;j++) { if(i==j) continue; if(x[i]==x[j]) k[cnt]=999999999; else { k[cnt]=(y[i]-y[j])/(x[i]-x[j]); } cnt++; } sort(k,k+cnt); num=2; for(p=1;p<cnt;p++) { if(k[p]==k[p-1]) { num++; if(max_num<num) max_num=num; } else { num=2; if(max_num<num) max_num=num; } } } cout<<max_num<<endl; } return 0; }
来源：https://www.cnblogs.com/w0w0/archive/2011/11/21/2256781.html