POJ-1118-简单暴力题

Lining Up 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 5 1 1 2 2 3 3 9 10 10 11 0 Sample Output 3 Source East Central North America 1994    这道题当初是搜搜索题的时候找到的。可是一看发现可以直接暴力所以也就没有搜索了...... #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; struct P { int x,y; }a[705]; int main() { int n; while(scanf("%d",&n)!=EOF&&n!=0) { for(int i=0;i<n;i++) scanf("%d%d",&a[i].x,&a[i].y); int ans=1; for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { int sum=2; for(int k=j+1;k<n;k++) { if((a[i].x-a[j].x)*(a[j].y-a[k].y)==(a[j].x-a[k].x)*(a[i].y-a[j].y)) sum++; } if(sum>ans) ans=sum; } } printf("%d\n",ans); }return 0; }