POJ-1118-简单暴力题

最新推荐文章于 2019-09-23 11:29:27 发布

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最新推荐文章于 2019-09-23 11:29:27 发布

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分类专栏：思维技巧文章标签： acm algorithm POJ brute force

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本文探讨了一种算法挑战：如何确定二维空间中最多数量的点位于同一直线上。通过暴力算法实现，输入为一系列坐标点，输出为一条直线上最多的点数。此问题源自East Central North America 1994的编程竞赛。

Lining Up

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 22296		Accepted: 7040

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

Sample Output

Source

East Central North America 1994

这道题当初是搜搜索题的时候找到的。可是一看发现可以直接暴力所以也就没有搜索了......

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct P
{
    int x,y;
}a[705];

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF&&n!=0)
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&a[i].x,&a[i].y);
        int ans=1;
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                int sum=2;
                for(int k=j+1;k<n;k++)
                {
                    if((a[i].x-a[j].x)*(a[j].y-a[k].y)==(a[j].x-a[k].x)*(a[i].y-a[j].y))
                        sum++;
                }
                if(sum>ans) ans=sum;
            }
        }
        printf("%d\n",ans);
    }return 0;
}