poj 1113 Wall

题目链接：http://poj.org/problem?id=1113

题目大意：给出点集和一个长度L，要求用最短长度的围墙把所有点集围住，并且围墙每一处距离所有点的距离最少为L，求围墙的长度。

解法：凸包+以L为半径的圆的周长。以题目中的图为例，两点之间的围墙长度之和正好就是凸包的长度，再加上每个点的拐角处（注意此处为弧，才能保证城墙距离点的距离最短）的长度。

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define inf 0x7fffffff
 #define exp 1e-10
 #define PI 3.141592654
 using namespace std;
 int n,L;
 struct Point
 {
     double x,y;
     Point (double x=,double y=):x(x),y(y){}
     friend bool operator < (Point a,Point b)
     {
         if (a.x!=b.x) return a.x<b.x;
         return a.y<b.y;
     }
 }an[+],bn[+];
 typedef Point Vector;
 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
 int dcmp(double x)
 {
     if (fabs(x)<exp) return ;
     else return x< ? - : ;
 }
 double cross(Vector A,Vector B)
 {
     return A.x*B.y-B.x*A.y;
 }
 int ConvexHull(Point *p,int n,Point *ch)
 {
     sort(p,p+n);
     int m=;
     for (int i= ;i<n ;i++)
     {
         while (m> && dcmp(cross(ch[m-]-ch[m-],p[i]-ch[m-]))<=) m--;
         ch[m++]=p[i];
     }
     int k=m;
     for (int i=n- ;i>= ;i--)
     {
         while (k>m && dcmp(cross(ch[k-]-ch[k-],p[i]-ch[k-]))<=) k--;
         ch[k++]=p[i];
     }
     ch[k++]=ch[];
     if (n>) k--;
     return k;
 }
 int main()
 {
     while (cin>>n>>L)
     {
         for (int i= ;i<n ;i++)
         scanf("%lf%lf",&an[i].x,&an[i].y);

         int k=ConvexHull(an,n,bn);
         double sum=;
         sum += *PI*L;
         for (int i= ;i<k ;i++)
         {
             sum += sqrt((bn[i].x-bn[i+].x)*(bn[i].x-bn[i+].x)+(bn[i].y-bn[i+].y)*(bn[i].y-bn[i+].y));
         }
         printf("%.0f\n",sum);
     }
     return ;
 }