POJ 1118 Lining Up && POJ 2606 Rabbit hunt 找出一条直线上的点的最大个数

http://poj.org/problem?id=2606

http://poj.org/problem?id=1118

这里是O(n^3)实现 水过！

Lining Up

Time Limit:2000MS

Memory Limit:32768K

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

/* Author : yan
 * Question : POJ 1118 Lining Up
 * Date && Time : Saturday, February 12 2011 06:44 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
#define  MAX 701
typedef struct
{
	int x,y;
}Point;
Point points[MAX];
int ans;

int main()
{
	freopen("input","r",stdin);
	int n,i,j,k;
	int tmp_max;
	while(scanf("%d",&n) && n)
	{
		for(i=0;i<n;i++) scanf("%d %d",&points[i].x,&points[i].y);
		ans=0;
		for(i=0;i<n;i++)
		{
			for(j=i+1;j<n;j++)
			{
				tmp_max=0;
				for(k=j+1;k<n;k++)
				{
					 if ((points[k].x - points[i].x)*(points[j].y - points[i].y) == (points[j].x - points[i].x)*(points[k].y - points[i].y))
						tmp_max++;
				}
				ans=ans>tmp_max?ans:tmp_max;
			}
		}
		printf("%d/n",ans+2);
	}
	return 0;
}

 

  Rabbit hunt
 
 

  
 
   

    
 
     
 Time Limit:1000MS
 
     
 
     
 Memory Limit:65536K
 
    
 
    
 
     

 
     

 
     

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

Sample Output
5

/* Author : yan
 * Question : POJ 2606 Rabbit hunt
 * Date && Time : Saturday, February 12 2011 07:28 PM
 * Compiler : gcc (Ubuntu 4.4.3-4ubuntu5) 4.4.3
*/
#include<stdio.h>
typedef struct
{
	int x,y;
}Point;
Point points[201];
int ans;

int main()
{
	//freopen("input","r",stdin);
	int n,i,j,k;
	scanf("%d",&n);
	for(i=0;i<n;i++) scanf("%d %d",&points[i].x,&points[i].y);
	int tmp_max;
	for(i=0;i<n;i++)
	{
		for(j=i+1;j<n;j++)
		{
			tmp_max=0;
			for(k=j+1;k<n;k++)
			{
				if( (points[i].x-points[j].x) * (points[i].y-points[k].y) == (points[i].x-points[k].x) * (points[i].y-points[j].y) ) tmp_max++;
			}
			ans=ans>tmp_max?ans:tmp_max;
		}
	}
	printf("%d",ans+2);
	return 0;
}