Codeforces 618C Constellation(简单几何题—叉积)

C. Constellation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Cat Noku has obtained a map of the night sky. On this map, he found a constellation with n stars numbered from 1 to n. For each i, thei-th star is located at coordinates (xi, yi). No two stars are located at the same position.

In the evening Noku is going to take a look at the night sky. He would like to find three distinct stars and form a triangle. The triangle must have positive area. In addition, all other stars must lie strictly outside of this triangle. He is having trouble finding the answer and would like your help. Your job is to find the indices of three stars that would form a triangle that satisfies all the conditions.

It is guaranteed that there is no line such that all stars lie on that line. It can be proven that if the previous condition is satisfied, there exists a solution to this problem.

Input

The first line of the input contains a single integer n (3 ≤ n ≤ 100 000).

Each of the next n lines contains two integers xi and yi ( - 109 ≤ xi, yi ≤ 109).

It is guaranteed that no two stars lie at the same point, and there does not exist a line such that all stars lie on that line.

Output

Print three distinct integers on a single line — the indices of the three points that form a triangle that satisfies the conditions stated in the problem.

If there are multiple possible answers, you may print any of them.

Sample test(s)

input

30 11 01 1

output

1 2 3

input

50 00 22 02 21 1

output

1 3 5

Note

In the first sample, we can print the three indices in any order.

In the second sample, we have the following picture.

Note that the triangle formed by starts 1, 4 and 3 doesn't satisfy the conditions stated in the problem, as point 5 is not strictly outside of this triangle (it lies on it's border).

题意：给出n个点，依次给出第i号点的坐标。  找一个三角形，且这个三角形内不能有其他点存在。输出这三个点的编号，答案有多组，输出任意一组。

题解：为了确定三角形内没有其他点。我们可以选择按照x或者y坐标大小排序。选择x或者y极值的三个点，如果这三个点在一条直线上，我们接着判断下一点是否能与前两个点构成三角形。

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL __int64struct node{LL x,y;int id;}dot[100010];int cmp(node a,node b)//我这里是按照y坐标从小到大排序的 {if(a.y==b.y)return a.x<b.x;return a.y<b.y;}int judge(node a){if((dot[1].x-a.x)*(dot[2].y-a.y)!=(dot[1].y-a.y)*(dot[2].x-a.x))//两条直线是否平行判断三点是否在一条直线上 return 1;return 0;}int main(){int n,i,j;while(scanf("%d",&n)!=EOF){for(i=1;i<=n;++i){scanf("%I64d%I64d",&dot[i].x,&dot[i].y);dot[i].id=i;}sort(dot+1,dot+n+1,cmp);for(i=3;i<=n;++i){if(judge(dot[i])){printf("%d %d %d\n",dot[1].id,dot[2].id,dot[i].id);break;}}}return 0;}