[CF598C] Nearest vectors

[CF598C] Nearest vectors - 计算几何

Description

有 n 个点，每个点表示原点到该点的向量，让你求出两个向量最小的夹角，输出向量的序号。

Solution

考虑暴力算出极角然后排序，用 atan2(y,x)

#include <bits/stdc++.h>
using namespace std;

#define int long long

const double pi = acos(-1);


signed main()
{
    struct point
    {
        long double x, y, a;
        int id;
        void gen_angle()
        {
            long double res = atan2(y, x);
            if (res < 0.0)
                res += 2 * pi;
            a = res;
        }

        long double angle() const
        {
            return a;
        }

        bool operator<(const point &rhs) const
        {
            return angle() < rhs.angle();
        }
    };

    ios::sync_with_stdio(false);

    int n;
    cin >> n;

    vector<point> p(n + 2);
    for (int i = 1; i <= n; i++)
        cin >> p[i].x >> p[i].y, p[i].gen_angle(), p[i].id = i;

    sort(&p[1], &p[n + 1]);
    p[n + 1] = p[1];

    long double min_value = 1e9;
    int min_pos = 0;

    for (int i = 1; i < n; i++)
    {
        long double now = abs(p[i + 1].angle() - p[i].angle());
        if (now < min_value)
        {
            min_value = now;
            min_pos = i;
        }
    }

    long double now = abs(-p[n].angle() + 2 * pi + p[1].angle());
    if (now < min_value)
    {
        min_value = now;
        min_pos = n;
    }

    int x = min_pos, y = min_pos + 1;

    cout << p[x].id << " " << p[y].id << endl;
}