hdoj1432Lining Up

最新推荐文章于 2018-09-26 17:14:36 发布

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本文探讨了一位飞行员面临的挑战：在一次直飞任务中，如何最大化覆盖预先设定的投放点。通过算法解决如何确定最大数量的点位于同一直线上的问题，并提供了一个高效的程序实现。

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Problem 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

The input consists of multiple test cases, and each case begins with a single positive integer on a line by itself indicating the number of points, followed by N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one blank and ended by a new-line character. No pair will occur twice in one test case.

Output

For each test case, the output consists of one integer representing the largest number of points that all lie on one line, one line per case.

Sample Input

Sample Output

代码：

#include<iostream>
#include<algorithm>
using namespace std;
struct stu{
	double x;
	double y;
}a[701];
int main()
{
	int i,j,k,t;
	while(scanf("%d",&t)!=EOF)
	{		
	for(i=1;i<=t;i++)
	{
	scanf("%lf%lf",&a[i].x,&a[i].y);
	}
	if(t==2)
	{
		printf("2\n");
	}
	else
	{
	int max=0;
	for(i=1;i<=t;i++)
	{
		for(j=i+1;j<=t;j++)
		{
			int ans=0;
			for(k=j+1;k<=t;k++)
			{				
				if((a[j].x-a[i].x)*(a[k].y-a[i].y)-(a[k].x-a[i].x)*(a[j].y-a[i].y)==0)
				{
					ans++;
				}
			}
			if(ans>max)
			{
				max=ans;
			}
		}
	}
	printf("%d\n",max+2);
}
}
	return 0;
}

思路：判断点是不是在一条线上。可以用交叉相乘相等计算，感觉也可以用斜率相等算，但写出来老是不对。