例题来自Hdu 1348

//几何求凸包模板
#include<bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 5;
const double PI = acos(-1);
#define all(a) a.begin(), a.end()
struct Point {
    int x, y;
    Point() {}
    Point(int x, int y) : x(x), y(y) {}
    Point operator - (const Point& ot) const {
        return Point(x - ot.x, y - ot.y);
    }
    int operator * (const Point& ot) const {//求叉积
        return x * ot.y - y * ot.x;
    }
    void scan() {  //读入一个点对
        scanf("%d%d", &x, &y);
    }
    bool operator < (const Point& rhs) const {
        return make_pair(x, y) < make_pair(rhs.x, rhs.y);
    }
    int sqr_len() const {
        return x * x + y * y;
    }
    double get_len() const {  //求(x,y)点到(0,0)点的距离
        return sqrtl(sqr_len());
    }
    int operator % (const Point& rhs) const {//向量数量积,向量a在向量b上的投影
        return x * rhs.x + y * rhs.y;
    }
};

struct Polygon {
    vector<Point> hull;
    Polygon(int n) {
        vector<Point> points(n);
        for (int i = 0; i < n; ++i) {
            points[i].scan();
        }
        sort(all(points));
        vector<Point> up, down;
        for (auto& pt : points) {
            while (up.size() > 1 && (up[up.size() - 2] - up.back()) * (pt - up.back()) >= 0) {
                up.pop_back();
            }
            up.push_back(pt);
            while (down.size() > 1 && (down[down.size() - 2] - down.back()) * (pt - down.back()) <= 0) {
                down.pop_back();
            }
            down.push_back(pt);
        }
        hull = up;
        for (int i = (int)down.size() - 2; i > 0; --i) {
            hull.push_back(down[i]);
        }
    }
    double sqr_len(int pos) const {  //求的是距离
        return (hull[(pos + 1) % hull.size()] - hull[pos]).get_len();
    }
    int size() const {
        return (int)hull.size();
    }
    double get_angle(int pos) const {
        auto a = hull[(pos + 1) % hull.size()] - hull[pos], b = hull[(pos + hull.size() - 1) % hull.size()] - hull[pos];
        double co = (a % b) / a.get_len() / b.get_len();//返回的是cos(a)的值
        return co;
    }
};

int n, L;
int main()
{
    int T, kase = 0;
    scanf("%d", &T);
    while (T--) {
        if (kase++)printf("\n");
        scanf("%d%d", &n, &L);
        Polygon a(n);
        double ans = 2 * PI*L;
        vector<double>angle;
        for (int i = 0; i < a.size(); i++) {
            ans += a.sqr_len(i);//第i段的长度
            angle.push_back(a.get_angle(i));//表示获得凸包a的第i个角a的cos值
        }
        printf("%.0lf\n", ans);
    }
}