USACO fc 构造凸包

　　本题的意思就是构造一个凸包然后求出凸包的周长。代码如下：

　　

/*
    ID: m1500293
    LANG: C++
    PROG: fc
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <iostream>
#include <vector>

using namespace std;
double eps = 1e-10;

struct P
{
    double x, y;
    P(double x=0, double y=0):x(x), y(y) {}
    double add(double a, double b){
        if(abs(a+b)<eps*(abs(a)+abs(b))) return 0;
        return a+b;
    }
    P operator + (P p){
        return P(add(x, p.x), add(y, p.y));
    }
    P operator - (P p){
        return P(add(x, -p.x), add(y, -p.y));
    }
    P operator *(double d){
        return P(x*d, y*d);
    }
    double dot(P p){                 //点积
        return add(x*p.x, y*p.y);
    }
    double det(P p){                 //差积
        return add(x*p.y, -y*p.x);
    }
}ps[10000 + 100];

double dist(P a, P b)
{
    return sqrt((b-a).dot(b-a));
}

bool cmp_x(const P& p, const P& q)
{
    if(p.x!=q.x) return p.x < q.x;
    return p.y < q.y;
}

vector<P> convex_hull(P *ps, int n)
{
    sort(ps, ps+n, cmp_x);
    int k = 0;          //凸包顶点数
    vector<P> qs(n*2);
    //构造凸包的下侧
    for(int i=0; i<n; i++)
    {
        while(k>1 && (qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0) k--;
        qs[k++] = ps[i];
    }
    //构造凸包的上侧
    for(int i=n-2,t=k; i>=0; i--)
    {
        while(k>t && (qs[k-1]-qs[k-2]).det(ps[i]-qs[k-1])<=0) k--;
        qs[k++] = ps[i];
    }
    qs.resize(k-1);
    return qs;
}

vector<P> hulls;
int main()
{
    freopen("fc.in", "r", stdin);
    freopen("fc.out", "w", stdout);
    int n;
    scanf("%d", &n);
    for(int i=0; i<n; i++)
    {
        double x, y;
        scanf("%lf%lf", &x, &y);
        ps[i] = P(x, y);
    }
    hulls = convex_hull(ps, n);

    double dis = 0;
    for(int i=1; i<hulls.size(); i++)
        dis += dist(hulls[i], hulls[i-1]);
    dis += dist(hulls[hulls.size()-1], hulls[0]);
    printf("%.2f
", dis);
    return 0;
}