[POJ]1118 Lining up

Lining Up
Time Limit: 2000MS

Memory Limit: 32768K
Total Submissions: 25990

Accepted: 8140
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

甚至不用搜索..O(n^3)也能过

//Writer:GhostCai && His Yellow Duck

#include
#include
#include
#include
using namespace std;
struct point{
    int x,y;
}a[705];

inline int read_d(){
    int i=0;
    char c;
    while(c=getchar(),c<'0'||c>'9');
    while(c>='0'&&c<='9'){
        i=i*10+c-'0';
        c=getchar();
    }
    return i;
}

inline bool cmp(point x,point y){
    return x.x < y.x ;
}
int main(){

    int n;
    while(1){
        scanf("%d",&n);
        if(n==0) return 0;
        for(int i=1;i<=n;i++){
            scanf("%d%d",&a[i].x ,&a[i].y );
      }
        int ans=1;
        sort(a+1,a+1+n,cmp);
        for(int i=1;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++;
                }
                ans=max(ans,sum);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}