Найти наименьший периметр треугольника

@dias amanov · Регистрация: 24.11.2018

Your a given coordinates of N points on the XY-plane.All points are different.Find the triangle with minimum possible perimeter.

Input:
First line contains N - total amoun of points (3 ≤ N ≤ 50). Then N lines are inputed.
Each line that describes coordinate contains two numbers.
Output:
Your task is to output three numbers that forms triangle with minimum perimeter.
Note that points are numbered from 1 to N.
If it's multiple solutions output any of them.
Input Output
5 1 2 4
0 0
1.3 0
-2 0.1
1 0
10 10

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C++

#include<iostream>
 
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
using namespace std;
#define pii pair<float, float>
#define mp make_pair
#define pb push_back
#define F first
#define S second
 
 
float dist(pii a, pii b) {
  return sqrt(pow(a.F - b.F, 2) + pow(a.S - b.S, 2));
}
int main() {
  int N;
  cin >> N;
  vector<pii> coords;
  for (int i = 0; i < N; i++) {
    float x, y;
    cin >> x >> y;
    coords.push_back(make_pair(x, y));
  }
  float minAB = dist(coords[0], coords[1]), minBC = dist(coords[0], coords[2]), minAC = dist(coords[1], coords[2]);
  int x = 0, y = 0, z = 0;
  float P = minAB + minBC + minAC;
  for (int i = 0; i < coords.size(); i++) {
    for (int j = 0; j < coords.size(); j++) {
      for (int k = 0; k < coords.size(); k++) {
        if (i != j && j != k && i != k) {
          minAB = dist(coords[i], coords[j]);
          minBC = dist(coords[j], coords[k]);
          minAC = dist(coords[i], coords[ k]);
          if (minAB + minBC + minAC < P) {
            P = minAB + minBC + minAC;
            x = i;
            y = j;
            z = k;
          }
        }
      }
    }
  }
  cout << x + 1 << " " << y + 1 << " " << z + 1;
  return 0;
}

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