85. Three Points On A Line

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本文介绍了一种用于判断二维平面上三个点是否共线的算法。通过输入多个测试用例，每个用例包含一系列点的坐标，算法将确定是否存在三点共线的情况。此算法在计算机图形学和几何计算中具有广泛应用。

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85. Three Points On A Line
时间限制 1000 ms 内存限制 65536 KB
题目描述
Given points on a 2D plane, judge whether there're three points that locate on the same line.

输入格式
The number of test cases appears in the first line of input.

Each test case begins with the number of points . The following lines describe the coordinates of each point, in accuracy of at most 3 decimals. Coordinates are ranged in .

输出格式
For each test case, output Yes if there're three points located on the same line, otherwise output No.

输入样例
2
3
0.0 0.0
1.0 1.0
2.0 2.0
3
0.001 -2.000
3.333 4.444
1.010 2.528
输出样例
Yes
No

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
struct node
{
    double x;
    double y;
}s[110];
int judge(node a, node b, node c)
{
   if((a.y - b.y)/(a.x - b.x) == (c.y - a.y)/(c.x - a.x))
        return 1;
    return 0;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; i++)
        {
            scanf("%lf%lf", &s[i].x,&s[i].y);
        }
        int flag = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = i+1; j < n; j++)
            {
                for(int k = j+1; k < n; k++)
                {
                    if(flag)
                        break;
                    flag = judge(s[i],s[j],s[k]);
                    
                }
            }
        }
        if(flag)
            printf("Yes\n");
        else
            printf("No\n");
    }
}