Lining Up Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status
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
5 1 1 2 2 3 3 9 10 10 11 0Sample Output
3
大意：求坐标点共线的最大数量
思路：任意找两个点作为一条边，然后判断其他点是否在在这条直线上
#include<cstdio> #include<algorithm> #include<cstring> #include<cmath> using namespace std; int n; int x[710],y[710]; bool judge(int x1,int y1,int x2,int y2,int x3,int y3) { if((y1-y3)*(x2-x3)==(y2-y3)*(x1-x3)) return 1; return 0; } int main() { while(scanf("%d",&n),n) { for(int i=1;i<=n;i++) scanf("%d%d",&x[i],&y[i]); int ans=0; for(int i=1;i<=n;i++) { for(int j=i+1;j<=n;j++) { int cnt=0; for(int k=1;k<=n;k++) { if(judge(x[i],y[i],x[j],y[j],x[k],y[k])) cnt++; } ans=max(ans,cnt); } } printf("%d\n",ans); } return 0; }
