平面凸包

本题是求凸包周长

#include <bits/stdc++.h>
#define rep(i, L, R) for (int i = L; i <= R; ++i)
using namespace std;
const double eps = 1e-8;
const double inf = 1e20;
const int N = 107;
inline int sgn(double x) {
    if (fabs(x) < eps) return 0;
    return x > 0 ? 1 : -1;
}
struct point {
    double x, y;
    point operator+(point b) { return {x + b.x, y + b.y}; };
    point operator-(point b) { return {x - b.x, y - b.y}; };
    point operator*(double k) { return {x * k, y * k}; };
    point operator/(double k) { return {x / k, y / k}; };
    bool operator==(point b) { return sgn(x - b.x) == 0 and sgn(y - b.x) == 0; }
    bool operator<(const point &b) const {
        return sgn(x - b.x) < 0 || sgn(x - b.x == 0) and sgn(y - b.y) == 0;
    }
};
typedef point pec;
double cross(pec a, pec b) { return a.x * b.y - a.y * b.x; }
double dis(point a, point b) { return hypot(a.x - b.x, a.y - b.y); }
double S(point *a, int n) {  //数组起始位置 多边形有多少个点
    double res = 0;
    for (int i = 0; i < n; ++i) res += cross(a[i], a[(i + 1) % n]);
    return res / 2;  // 面积有正负
}
int convexHull(point *a, int n, point *res) {
    sort(a, a + n);
    n = unique(a, a + n) - a;
    int p = 0;
    // 求下凸包 如果点是右拐的 回退 不收纳
    for (int i = 0; i < n; ++i) {
        while (p > 1 and
               sgn(cross(res[p - 1] - res[p - 2], a[i] - res[p - 2])) <= 0)
            --p;
        res[p++] = a[i];
    }
    // 求上凸包
    int j = p;
    for (int i = n - 2; i >= 0; --i) {
        while (p > j and
               sgn(cross(res[p - 1] - res[p - 2], a[i] - res[p - 2])) <= 0)
            --p;
        res[p++] = a[i];
    }
    if (n > 1) --p;
    return p; // 凸包顶点数
}
point a[N], res[N];
int main() {
    int n;
    while (~scanf("%d", &n) && n) {
        rep(i, 1, n) scanf("%lf%lf", &a[i].x, &a[i].y);
        int v = convexHull(a + 1, n, res);
        double ans = 0;
        if (v == 1)
            ans = 0;
        else if (v == 2)
            ans = dis(res[0], res[1]);
        else  // 凸包周长
            for (int i = 0; i < v; ++i) ans += dis(res[i], res[(i + 1) % v]);
        printf("%.2lf\n", ans);
    }
    return 0;
}