POJ 1113&&HDU 1348

题意：凸包周长+一个完整的圆周长。因为走一圈，经过拐点时，所形成的扇形的内角和是360度，故一个完整的圆。

模板题，之前写的Graham模板不对，WR了很多发。。。。POJ上的AC代码

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<set>
 #include<stdio.h>
 #include<stdlib.h>
 #include<math.h>
 #define clc(a,b) memset(a,b,sizeof(a))
 #define eps 1e-8
 #define PI acos(-1.0)
 typedef long long LL;
 const int mod=;
 const int inf=0x3f3f3f3f;
 const int MAXN=;
 using namespace std;

 const int N = ;

 int top;
 struct point
 {
     double x;
     double y;
 } p[N], Stack[N];

 double dis(point A, point B)
 {
     return sqrt((A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y));
 }

 double crossProd(point A, point B, point C)
 {
     return (B.x-A.x)*(C.y-A.y) - (B.y-A.y)*(C.x-A.x);
 }

 int cmp(const void *a, const void *b)
 {
     point *c = (point *)a;
     point *d = (point *)b;
     double k = crossProd(p[], *c, *d);
     if (k< || !k && dis(p[], *c)>dis(p[], *d))
     return ;
     return -;
 }

 void Graham(int n)
 {
     double x = p[].x;
     double y = p[].y;
     int mi = ;
     for (int i=; i<n; ++i)
     {
         if (p[i].x<x || (p[i].x==x && p[i].y<y))
         {
             x = p[i].x;
             y = p[i].y;
             mi = i;
         }
     }
     point tmp = p[mi];
     p[mi] = p[];
     p[] = tmp;
     qsort(p+, n-, sizeof(point), cmp);
     p[n] = p[];
     Stack[] = p[];
     Stack[] = p[];
     Stack[] = p[];
     top = ;
     for (int i=; i<=n; ++i)
     {
         while (crossProd(Stack[top-], Stack[top], p[i])<= && top>=) --top;
         Stack[++top] = p[i];
     }
 }

 int main()
 {
     int n, l;
     while (scanf("%d%d",  &n, &l) != EOF)
     {
             for (int i=; i<n; ++i)
             {
                 scanf ("%lf%lf", &p[i].x, &p[i].y);
             }
             Graham(n);
             double ans=;
             for(int i=;i<top;i++)
             {
                 ans+=dis(Stack[i], Stack[i+]);
             }
             ans += PI * (l + l);
             printf ("%d\n", (int)(ans+0.5));
     }
     return ;
 }