Wunder Fund Round 2016 (Div. 1 + Div. 2 combined)--C. Constellation

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

3
0 1
1 0
1 1

output

1 2 3

input

5
0 0
0 2
2 0
2 2
1 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).

大体题意：

给你很多点，寻找三个点使得他们围成的三角形内部没有其他点，三条线上也不能有！

分析：

自己一个方法没过，想要用极点排序做，先排出角来，在x，y排序，其实不用，直接按照先x后y的顺序排序就行了，就能保证相邻的的三个点三角形没有其他点。在取第一第二两个点固定，扫描第三个点，判断是否构成三角形！

这里判断方法用向量做，判断两向量是否平行，

需要注意的是坐标最好用long long !

代码如下：

#include<algorithm>
#include<iostream>
using namespace std;
const int maxn  = 100000 + 10;
typedef long long ll;
struct point{
    ll x,y,id;
    bool operator < (const point &a)const{
        return x < a.x || (x == a.x && y < a.y);
    }
}po[maxn];
bool judge(int i){
    ll x1 = po[0].x,y1=po[0].y,x2=po[1].x,y2=po[1].y,x3=po[i].x,y3=po[i].y;
    if ((x2-x1)*(y3-y2) != (x3-x2)*(y2-y1))return true;
    return false;
}
int main()
{
    int n;
    while(cin >> n && n){
        for (int i = 0; i < n; ++i){cin >> po[i].x >> po[i].y;po[i].id=i+1;}
        sort(po,po+n);
        for (int i = 2; i < n; ++i){if (judge(i)){cout << po[0].id << " " << po[1].id << " " << po[i].id <<endl;break;}}
    }
    return 0;
}