本文主要是介绍BUPT OJ85 Three Points On A Line，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
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 T(1≤T≤10) appears in the first line of input.
Each test case begins with the number of points N(1≤N≤100) . The following N lines describe the coordinates (xi,yi) of each point, in accuracy of at most 3 decimals. Coordinates are ranged in [−104,104] .
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暴力
/*
USER_ID: test#birdstorm
PROBLEM: 85
SUBMISSION_TIME: 2014-03-05 21:08:58
*/
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define For(i,m,n) for(i=m;i<n;i++)
#define MAXN 105main()
{int t, i, j, k, m, n, flag;double a[MAXN], b[MAXN];scanf("%d",&t);while(t--){flag=0;scanf("%d",&n);scanf("%lf%lf%lf%lf",&a[0],&b[0],&a[1],&b[1]);For(m,2,n){scanf("%lf%lf",&a[m],&b[m]);for(i=0;i<m-1&&!flag;i++) For(j,i+1,m)if(fabs(a[i]*b[j]-a[j]*b[i]+a[j]*b[m]-a[m]*b[j]+a[m]*b[i]-a[i]*b[m])<=1e-6) flag=1;}puts(flag?"Yes":"No");}return 0;
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