Poj 1113 Wall (凸包Graham水平序)

╰+哭是因爲堅強的太久メ 2023-01-20 11:57 22阅读 0赞

以前做过的题，现在重新写下。

以前的解题报告：http://blog.csdn.net/whyorwhnt/article/details/8316442

#include <cmath>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const double PI=acos(-1.0);
    const double eps=1e-8;
    const int N=1005;
    
    struct Point  //点,向量
    {
        double x,y;
        Point(){}
        Point(double _x,double _y)
        {x=_x;y=_y;}
        void Get ()
        {scanf("%lf%lf",&x,&y);}
    
        Point operator-(const Point &b)const
        {return Point(x-b.x,y-b.y);}
    
        Point operator+(const Point &b)
        {return Point(x+b.x,y+b.y);}
    
        Point operator*(const double k)  //数乘
        {return Point(x*k,y*k);}
    
        double operator*(const Point a)  //点乘
        {return x*a.x+y*a.y;}
    
        double operator^(const Point a)  //叉乘
        {return x*a.y-y*a.x;}
    
        int DB (double dx)  //判0函数
        {
            if(fabs(dx)<eps) return 0;
            return dx>0?1:-1;
        }
        //调用点a的该函数
        //返回1点a在向量bc的左侧
        //返回-1点a在向量bc的右侧
        //返回0点a在向量bc这条直线上
        int Cross (Point b,Point c)
        {return DB((b.x-x)*(c.y-y)-(c.x-x)*(b.y-y));}
    
        bool operator < (const Point &b)const //用于排序
        {
            if (fabs(y-b.y)<eps) return x-b.x<0;
            return y-b.y<0;
        }
    
        double Dis (Point ch)
        {return sqrt((x-ch.x)*(x-ch.x)+(y-ch.y)*(y-ch.y));}
    }pt[N],ch[N];
    
    int n,len;
    
    //求凸包函数
    //pt:所有的点，n个,pt[0]到pt[n-1]
    //ch:求完的凸包中的点，len个,q[0]到q[len-1]
    void Graham (Point pt[],Point ch[],int &len,int n)      //只保存凸包顶点
    {
    	int i,top=1;
    	sort(pt,pt+n);
        ch[0]=pt[0];
        ch[1]=pt[1];
        for (i=2;i<n;i++)
        {
            while (top>0 && ch[top-1].Cross(ch[top],pt[i]) <= 0)
                top--;
            ch[++top]=pt[i];
        }
        int temp=top;
        for (i=n-2;i>=0;i--)
        {
            while (top>temp && ch[top-1].Cross(ch[top],pt[i]) <= 0)
                top--;
            ch[++top]=pt[i];
        }
        len=top;
    }
    
    int main ()
    {
        int L,i,n,len;
        double ans=0;
        scanf("%d%d",&n,&L);
        for (i=0;i<n;i++)
            pt[i].Get();
        Graham (pt,ch,len,n);
        for (i=0;i<len;i++)
            ans+=ch[i].Dis(ch[(i+1)%len]);
        ans+=PI*2*L;
        printf("%.0lf\n",ans);
        return 0;
    }