C - 极角 CodeForces - 598C （数学）-优快云博客

本文介绍了一个CodeForces上的算法竞赛题目——寻找平面上多个向量间最小非定向角度的解决方案。通过计算各向量与x轴正方向的角度并排序，比较相邻向量角度差来找出最小角度对应的向量对。

C - 极角

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 nin 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.

Examples

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：给出n个点的坐标，这些点分别和（0，0）相连，问这些线之间角度（【0，PI】）最小的是哪些，输出其中一对的编号。

求出这些线和x正半轴之间的角度，然后排序，再将相邻的两个角度相减，记录最小的

PS：这题高精度用long double；有可能出现【0，PI】之外的，要转到【0，PI】。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <queue>
#include <set>
#include <map>
#define loopa(i,a,b,d) for(int i=a;i<b;i+=d)
#define loops(i,a,b,d) for(int i=a;i>b;i-=d)

typedef long long LL;
typedef long double LD;
using namespace std;
const int maxn=100000+1;
const LD PI = acos(-1.0);
struct node
{
    LD d;
    int x;
};
node a[maxn];
bool cmp(node a,node b)
{
    return a.d<b.d;
}
int main()
{
    int n;

    while(~scanf("%d",&n))
    {
        loopa(i,0,n,1)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            a[i].d=atan2((LD)y,(LD)x);
            a[i].x=i+1;
        }
        sort(a,a+n,cmp);
        LD su=100;
        int ans1,ans2;
        loopa(i,0,n-1,1)
        {
            LD t=a[i+1].d-a[i].d;

            if(t<=su)
            {
                ans1=a[i+1].x;
                ans2=a[i].x;
                su=a[i+1].d-a[i].d;
            }
        }
        LD t=a[n-1].d-a[0].d;
        if(t<0)t+=2*PI;
        if(t>PI)t=2*PI-t;
        if(t<=su)
        {
            ans1=a[n-1].x;
            ans2=a[0].x;
            su=a[n-1].d-a[0].d;
        }
        printf("%d %d\n",ans1,ans2);
    }
    return 0;
}

转载于:https://www.cnblogs.com/107acm/p/9428333.html