POJ 1118 Lining Up
2016-09-01 16:44
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Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %lluDescription"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!InputInput 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.Outputoutput 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 0Sample Output
3 看注释
#include <cstdio> #include <cstring> #include <cstdlib> #include <cmath> #include <algorithm> using namespace std; struct node { int x,y; }a[705]; int main() { int n,i,ans,j,f,m; while(scanf("%d",&n)!=EOF&&n!=0) { for(i=1;i<=n;i++) { scanf("%d%d",&a[i].x,&a[i].y); } ans=2; for(i=1;i<=n;i++) { for(j=i+1;j<=n;j++) { //选择两点连成直线 f=2; for(m=j+1;m<=n;m++)//节约时间,避免重复验证 { if((a[m].x-a[i].x)*(a[m].y-a[j].y)==(a[m].x-a[j].x)*(a[m].y-a[i].y)) f++; } if(f>ans) ans=f; } } printf("%d\n",ans); } return 0; }
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