Educational Codeforces Round 1 C.Nearest vectors(排序)

Educational Codeforces Round 1C：http://codeforces.com/contest/598/problem/C

C. Nearest vectors

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given the set of vectors on the plane, each of them starting at the origin. Your task is to find a pair of vectors with the minimal non-oriented angle between them.

Non-oriented angle is non-negative value, minimal between clockwise and counterclockwise direction angles. Non-oriented angle is always between 0 and π. For example, opposite directions vectors have angle equals to π.

Input

First line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of vectors.

The i-th of the following n lines contains two integers xi and yi (|x|, |y| ≤ 10 000, x2 + y2 > 0) — the coordinates of the i-th vector. Vectors are numbered from 1 to n in order of appearing in the input. It is guaranteed that no two vectors in the input share the same direction (but they still can have opposite directions).

Output

Print two integer numbers a and b (a ≠ b) — a pair of indices of vectors with the minimal non-oriented angle. You can print the numbers in any order. If there are many possible answers, print any.

Sample test(s)

input

4
-1 0
0 -1
1 0
1 1

output

3 4

input

6
-1 0
0 -1
1 0
1 1
-4 -5
-4 -6

output

6 5

题目大意：给一堆向量，找出角度差最小的两个向量

大致思路：按照角度递增排序，枚举相邻两个角度，注意还需要比较最小角和最大角的角度差。

坑点是double精度不够，必须用long double，发现scanf();无法读入long double类型，%Lf与%llf均出错

#include <iostream>
#include <cmath>
#include <algorithm>

using namespace std;

int n,i,p,q;
long double x,y,mn,t;
const long double PI=acos((long double) (-1));

struct Node {
    long double d;
    int id;
    bool operator <(const Node& a) const {//按照角度递增排序
        return d<a.d;
    }
}v[100005];

int main() {
	while(cin>>n) {
        for(i=1;i<=n;++i) {
            cin>>x>>y;
            v[i].id=i;
            v[i].d=atan2(y,x);//atan2(y,x)能求出PI/2等tan值不存在的角
        }
        sort(v+1,v+n+1);
        mn=2*PI-v[n].d+v[1].d;//初始化最小角度差为最大角与最小角间的角度差
        p=v[1].id;
        q=v[n].id;
        for(i=1;i<n;++i) {
            t=v[i+1].d-v[i].d;
            t=min(t,2*PI-t);//由于没有方向，所以夹角一定是t和t的补角中较小的
            if(t<mn) {
                mn=t;
                p=v[i].id;
                q=v[i+1].id;
            }
        }
        cout<<p<<" "<<q<<"\n";
	}
	return 0;
}