POJ 2606 Rabbit hunt

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本文介绍了一种解决猎兔问题的算法，通过判断多个点是否共线来找出可被一击射中的最多兔子数量。输入为平面上带有整数坐标的一系列点，输出则为这些点中位于同一直线上的最大点数。

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Rabbit hunt

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8137		Accepted: 4058

Description

A good hunter kills two rabbits with one shot. Of course, it can be easily done since for any two points we can always draw a line containing the both. But killing three or more rabbits in one shot is much more difficult task. To be the best hunter in the world one should be able to kill the maximal possible number of rabbits. Assume that rabbit is a point on the plane with integer x and y coordinates. Having a set of rabbits you are to find the largest number K of rabbits that can be killed with single shot, i.e. maximum number of points lying exactly on the same line. No two rabbits sit at one point.

Input

An input contains an integer N (2<=N<=200) specifying the number of rabbits. Each of the next N lines in the input contains the x coordinate and the y coordinate (in this order) separated by a space (-1000<=x,y<=1000).

Output

The output contains the maximal number K of rabbits situated in one line.

Sample Input

Sample Output5

思想：两个点连成直线，再判断第三个点是否在该直线上

#include <iostream>
#include <cstdio>
using namespace std;
#define N 1000
#define up(i,a,b) for(int i=a;i<=b;i++)//定义一个循环
struct node{
    int x,y;
};

int main(int argc, const char * argv[]) {
    int max=0;
    int n;
    cin>>n;
    node t[n];
    int i=0;
    up(i,0,n-1)
    scanf("%d %d",&t[i].x,&t[i].y);
    for(i=0;i<n;i++)//三个点里的第一个点
        for(int j=i+1;j<n;j++)//三个点里的第二个点
        {
            int sum=0;
            for(int k=j+1;k<n;k++)//三个点里的第三个点
            {
                if((t[i].y-t[j].y)*(t[i].x-t[k].x)==(t[i].y-t[k].y)*(t[i].x-t[j].x))//斜率相等
                    sum++;
            }
            if(sum>max)
                max=sum;
        }
    cout<<max+2<<endl;//注意要换行
    return 0;
}