URAL 2067 Friends and Berries

There is a group of n children. According to a proverb, every man to his own taste. So the children value strawberries and raspberries differently. Let’s say that
i-th child rates his attachment to strawberry as s i and his attachment to raspberry as
r i.

According to another proverb, opposites attract. Surprisingly, those children become friends whose tastes differ.

Let’s define friendliness between two children
v, u as: p( v, u) = sqrt((
s v − s u) 2 + (
r v − r
u) 2)

The friendliness between three children v,
u, w is the half the sum of pairwise friendlinesses: p(
v, u, w) = ( p( v, u) +
p( v, w) + p( u, w)) / 2

The best friends are that pair of children v,
u for which v ≠ u and p( v, u) ≥
p( v, u, w) for every child w. Your goal is to find all pairs of the best friends.

Input

In the first line there is one integer n — the amount of children (2 ≤
n ≤ 2 · 10 5).

In the next n lines there are two integers in each line —
s i and r
i (−10 8 ≤
s i, r i ≤ 10 8).

It is guaranteed that for every two children their tastes differ. In other words, if
v ≠ u then s v ≠
s u or r
v ≠ r u.

Output

Output the number of pairs of best friends in the first line.

Then output those pairs. Each pair should be printed on a separate line. One pair is two numbers — the indices of children in this pair. Children are numbered in the order of input starting from 1. You can output pairs in any
order. You can output indices of the pair in any order.

It is guaranteed that the amount of pairs doesn’t exceed 105.

Example

input	output
2 2 3 7 6	1 1 2
3 5 5 2 -4 -4 2	0

Hint

题目中那两个公式的意思就是如果两个点与其他任意一个点都共线并在同一边上，就存在最好的朋友，否则输出0先排序，最远的两个点即为所求。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
struct node
{
int x,y;
int num;
}a[300005];
bool cmp(node a,node b)
{
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
int main()
{
int n;
scanf("%d",&n);
int i;
for(i=0;i<=n-1;i++)
{
scanf("%d%d",&a[i].x,&a[i].y);
a[i].num=i+1;
}
sort(a,a+n,cmp);
int flag=0;
for(i=0;i<=n-1;i++)
{
if(abs(a[i].x-a[0].x)*abs(a[i].y-a[n-1].y)!=abs(a[i].x-a[n-1].x)*abs(a[i].y-a[0].y))
{
flag=1;
break;
}
}
if(!flag)
{
printf("1\n");
printf("%d %d\n",a[0].num,a[n-1].num);
}
else printf("0\n");
return 0;
}

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