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 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 (x i ,y i ) of each point, in accuracy of at most 3 decimals. Coordinates are ranged in [−10 4 ,10 4 ] .

输出格式

For each test case, output Yes if there're three points located on the same line, otherwise outputNo.

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

三点共线，枚举所有的3点，求斜率，一样则存在，做好标记然后退出枚举即可。还有一种做法是求出一组后，把计算结果存起来，减少计算量。

#include<stdio.h>
int main(){int t,n,i,j,k;double x[105],y[105];for(scanf("%d",&t);t--;){scanf("%d",&n);for(i=0;i<n;i++)scanf("%lf%lf",x+i,y+i);int sig=0;for(i=0;i<n&&!sig;i++)for(j=i+1;j<n&&!sig;j++){for(k=j+1;k<n&&!sig;k++){if((y[k]-y[i])/(x[k]-x[i])==(y[j]-y[i])/(x[j]-x[i]))sig=1;}}if(sig)printf("Yes\n");else printf("No\n");}return 0;
}

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北邮OJ 85. Three Points On A Line