凸包

水一下。。构造凸包求周长。。

#include 
    
     #define INF 0x3f3f3f3f#define full(a, b) memset(a, b, sizeof a)#define FAST_IO ios::sync_with_stdio(false)using namespace std;typedef long long LL;inline int lowbit(int x){ return x & (-x); }inline int read(){    int ret = 0, w = 0; char ch = 0;    while(!isdigit(ch)){        w |= ch == '-', ch = getchar();    }    while(isdigit(ch)){        ret = (ret << 3) + (ret << 1) + (ch ^ 48);        ch = getchar();    }    return w ? -ret : ret;}inline int lcm(int a, int b){ return a / __gcd(a, b) * b; }template 
     
      inline A fpow(A x, B p, C lyd){    A ans = 1;    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;    return ans;}#define eps 1e-6;struct Point{    double x, y;    Point(double x, double y): x(x), y(y){}    Point operator + (Point &a){        return Point(x + a.x, y + a.y);    }    Point operator - (Point &a){        return Point(x - a.x, y - a.y);    }    bool operator < (const Point &a) const {        if(x == a.x) return y < a.y;        return x < a.x;    }    bool operator == (Point &a){        if(x == a.x && y == a.y) return true;        return false;    }    double abs(){        return sqrt(x * x + y * y);    }};typedef Point Vector;vector
      
        p, u, l;int n;double cross(Vector a, Vector b){    return a.x * b.y - a.y * b.x;}bool isClock(Point a, Point b, Point c){    Vector x = b - a;    Vector y = c - a;    return cross(x, y) < eps;}double dis(Point a, Point b){    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));;}int main(){    while(~scanf("%d", &n) && n){        p.clear(), u.clear(), l.clear();        for(int i = 1; i <= n; i ++){            double x, y;            scanf("%lf%lf", &x, &y);            p.push_back(Point(x, y));        }        if(n == 1){            printf("0.00\n");            continue;        }        if(n == 2){            printf("%.2lf\n", dis(p[0], p[1]));            continue;        }        sort(p.begin(), p.end());        u.push_back(p[0]), u.push_back(p[1]);        for(int i = 2; i < p.size(); i ++){            for(int n = u.size(); n >= 2 && (!isClock(u[n - 2], u[n - 1], p[i])); n --)                u.pop_back();            u.push_back(p[i]);        }        l.push_back(p[p.size() - 1]), l.push_back(p[p.size() - 2]);        for(int i = p.size() - 3; i >= 0; i --){            for(int n = l.size(); n >= 2 && (!isClock(l[n - 2], l[n - 1], p[i])); n --)                l.pop_back();            l.push_back(p[i]);        }        for(int i = 1; i < u.size() - 1; i ++){            l.push_back(u[i]);        }        double ans = 0;        for(int i = 1; i < l.size(); i ++){            ans += dis(l[i], l[i - 1]);        }        ans += dis(l[0], l[l.size() - 1]);        printf("%.2lf\n", ans);    }    return 0;}