Timus 1052 Rabbit Hunt



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 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 (3 ≤  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 (−2000 ≤  x,  y ≤ 2000).

Output

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

Sample Input

6                           5
7 122
8 139
9 156
10 173
11 190
-100 1

题意+思路：在一个坐标格中，从给出的每个坐标点出发，与给出的其他点分别连成直线。题中说每一枪打到最多的兔子，就是找到一条直线，该条直线上含的坐标点最多，就相当于找连线斜率相同次数最多的相同次数。

注意：从一个点出发，最后次数加这个点的兔子才是在这条直线上兔子的总数。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;

const int maxn=505;
double a[maxn];

struct point
{
	double x,y;
}arr[maxn];

int main()
{
	int n;
	cin>>n;
	int k,max;
	for(int i=0;i<n;i++)
	{
		cin>>arr[i].x>>arr[i].y;
	}
	max=-1;
	for(int i=0;i<n;i++)
	{
		k=0;
		map<double,int>mp;
		for(int j=i+1;j<n;j++)
		{
			a[k]=(arr[i].y-arr[j].y)/(arr[i].x-arr[j].x);
			mp[a[k]]++;
			k++;
		}
		map<double,int>::iterator p1;
		for(p1=mp.begin();p1!=mp.end();p1++)
		{
			if(p1->second>max) max=p1->second;
		}
	}
	printf("%d\n",max+1);
	return 0;
}