POJ - 1118 - Lining Up （暴力枚举）

Lining Up Time Limit: 2000MS   Memory Limit: 32768K Total Submissions: 25940   Accepted: 8125 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

题意：一个平面上有n个点，问最多有几个点在同一条直线上

因为n并没有很大，所以考虑的就是把所有点都跑一遍，计算斜率可能会导致精度误差而且还需要另外考虑斜率不存在的问题，所以就用

 y1 / x1 = y2 / x2  =>  y1 * x2 = y2 * x1 来判断三点是否在同一直线

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;

const int maxn = 700+10;

int n,ans,vis[maxn];

struct node{int x,y;}p[maxn];

int main()
{
    while(scanf("%d",&n)&&n){
        for(int i=0;i<n;i++) scanf("%d%d",&p[i].x,&p[i].y);
        
        ans = 0;
        
        for(int i=0;i<n-1;i++){
            
            memset(vis,0,sizeof vis);
            
            for(int j=i+1;j<n;j++){
                
                if(vis[j]) continue;
                
                int dx = p[i].x - p[j].x,dy = p[i].y - p[j].y,cnt = 2;
                
                for(int k=j+1;k<n;k++){
                    
                    int dxx = p[i].x - p[k].x,dyy = p[i].y - p[k].y;
                    
                    if(dxx*dy==dx*dyy){
                        vis[k] = 1;
                        cnt++;
                    }
                }
                if(cnt>ans) ans = cnt;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}