bzoj 1670 Building the Moat

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凸包裸题。。。之前我从来没涉及过计算几何，现在觉得还是要学一学，至少可以写个暴力23333

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>using namespace std;#define N 5010#define epx 1e-8const double pi=atan(1.0)*4;int n,m,top;double Ans;struct Point{    double x,y;    Point operator + (const Point &a) const{        return (Point){x+a.x,y+a.y};    }    Point operator - (const Point &a) const{        return (Point){x-a.x,y-a.y};    }    double operator * (const Point &a) const{        return x*a.y-y*a.x;    }    bool operator == (const Point &a) const{        return fabs(x-a.x)<epx&&fabs(y-a.y)<epx;    }    bool operator != (const Point &a) const{        return !((Point){x,y}==a);    }}p[N],S[N];double sqr(double x){return x*x;};bool Cmp(const Point &a,const Point &b){    if(fabs((b-p[1])*(a-p[1]))<epx)        return a.y<b.y;    else return (a-p[1])*(b-p[1])>0;}inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}    return x*f;}void Graham(){    int t=1;    for(int i=2;i<=n;i++)        if(p[i].y<p[t].y||(p[i].y==p[t].y&&p[i].x<p[t].x))            t=i;    swap(p[1],p[t]);    sort(p+2,p+n+1,Cmp);    S[++top]=p[1];S[++top]=p[2];    for(int i=3;i<=n;i++)    {        while((S[top]-S[top-1])*(p[i]-S[top-1])<=epx&&top>1)            top--;        S[++top]=p[i];    }    S[top+1]=p[1];    for(int i=1;i<=top;i++)        Ans+=sqrt(sqr(S[i].x-S[i+1].x)+sqr(S[i].y-S[i+1].y));}int main(){    n=read();    for(int i=1;i<=n;i++)        p[i].x=read(),p[i].y=read();    Graham();    printf("%.2lf\n",Ans);    return 0;}

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