Lining Up-多点共线问题-优快云博客

本文探讨了如何通过编程计算给定点集中最多共线的点数。详细解释了判断三点共线的方法，并提供了输入输出样例及代码实现，旨在帮助读者理解和解决相关问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Lining Up

``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 begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input consists of N pairs of integers, where 1 < N < 700. Each pair of integers is separated by one blank and ended by a new-line character. The list of pairs is ended with an end-of-file character. No pair will occur twice.

Output

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.
The output consists of one integer representing the largest number of points that all lie on one line.

Sample Input

Sample Output

题意:给出几组点的坐标,求最多有几点共线.

思路:判断三点共线:向量AB=(x1,y1),向量AC=(x2,y2),若向量AB,AC共线,且交于同一点A,则A,B,C三点共线

那么如何判断两向量共线呢?x1*y2-x2*y1=0 <==> 向量AB//AC

坑爹的输入啊,runtime 了n次,最后看别人的代码才发现输入错了.花了我大把大把的时光啊

#include<stdio.h>
#include<string.h>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
struct node
{
    double x,y;
} s[800];
char str[20];
int main()
{
    int n,i,l,sum,cnt,j,k;
    double x1,x2,y1,y2;
    scanf("%d\n",&n);
    while(n--)
    {
        l=0;
        while(gets(str))
        {
            if(!strlen(str))
                break;
            sscanf(str,"%lf %lf",&s[l].x,&s[l].y);
            l++;
        }
        sum=0;
        for(i=0; i<l; i++)
        {
            for(j=i+1; j<l; j++)
            {
                x1=s[i].x-s[j].x;
                y1=s[i].y-s[j].y;
                cnt=2;
                for(k=j+1; k<l; k++)
                {
                    x2=s[i].x-s[k].x;
                    y2=s[i].y-s[k].y;
                    if(x1*y2-x2*y1==0)
                    {
                        cnt++;
                    }
                }
                if(cnt>sum)
                    sum=cnt;
            }
        }
        printf("%d\n",sum);
        if(n)
            putchar('\n');
    }
}