Wunder Fund Round 2016 (Div. 1 + Div. 2 combined) CF618C Constellation

C. Constellation
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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, the i-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

题意：
给出一系列点，求出一个三角形，要求其他的点不在三角形内部或者三角形的线段边上，换句话说就是其他点必须与此三角形相离。

#include <bits/stdc++.h>using namespace std;pair<long long,long long>p[110000];map<pair<long long,long long>,long long> mp;int main(){    int n;    while(cin >> n)    {        for (long long i = 0; i < n; i++)        {            cin >> p[i].first >> p[i].second;            mp[p[i]] = i + 1;        }        sort(p,p + n);        int flag = 2;        while((p[0].second-p[1].second)*(p[0].first-p[flag].first)==(p[0].first-p[1].first)*(p[0].second-p[flag].second))            ++flag;        cout << mp[p[0]] << " " << mp[p[1]] << " " << mp[p[flag]]<<endl;    }    return 0;}