写出蛮力法求解凸包问题的c++代码

以下是使用蛮力法求解凸包问题的C++代码：

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

struct Point {
    int x, y;
};

// 求解向量 OA 和向量 OB 的叉积
int cross(const Point& O, const Point& A, const Point& B) {
    return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
}

// 求解凸包问题
vector<Point> convexHull(vector<Point>& points) {
    int n = points.size();
    if (n <= 1) return points;

    sort(points.begin(), points.end(), [](Point& a, Point& b) {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    });

    vector<Point> hull;
    for (int i = 0; i < n; ++i) {
        while (hull.size() >= 2 &&
               cross(hull[hull.size()-2], hull[hull.size()-1], points[i]) < 0)
            hull.pop_back();
        hull.push_back(points[i]);
    }

    // 反转向量，从右到左构建下半部分的凸包
    reverse(points.begin(), points.end());
    hull.pop_back();
    for (int i = 0; i < n; ++i) {
        while (hull.size() >= 2 &&
               cross(hull[hull.size()-2], hull[hull.size()-1], points[i]) < 0)
            hull.pop_back();
        hull.push_back(points[i]);
    }

    // 删除重复的点
    sort(hull.begin(), hull.end(), [](Point& a, Point& b) {
        return a.x < b.x || (a.x == b.x && a.y < b.y);
    });
    hull.erase(unique(hull.begin(), hull.end()), hull.end());

    return hull;
}

int main() {
    vector<Point> points = {{0, 0}, {1, 1}, {2, 2}, {3, 3}, {3, 1}, {2, 1}};
    vector<Point> hull = convexHull(points);

    for (Point p : hull)
        cout << "(" << p.x << ", " << p.y << ")" << endl;

    return 0;
}

该代码的时间复杂度为 $O(n^2)$，其中 $n$ 是点集中点的数量。