题链：
http://poj.org/problem?id=1113
题解：
计算几何，凸包
题意：修一圈围墙把给出的点包围起来，且被包围的点距离围墙的距离不能小于L，求围墙最短为多少。
答案其实就是等于N个点的凸包的周长+半径为L的圆的周长。
代码：
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 1050
using namespace std;
const double eps=1e-8,Pi=acos(-1.0);
int sign(double x){
if(fabs(x)<=eps) return 0;
return x<0?-1:1;
}
struct Point{
double x,y;
Point(double _x=0,double _y=0):x(_x),y(_y){}
void Read(){scanf("%lf%lf",&x,&y);}
};
typedef Point Vector;
bool operator < (const Point A,const Point B){return sign(A.x-B.x)<0||(sign(A.x-B.x)==0&&sign(A.y-B.y)<0);}
bool operator == (const Point A,const Point B){return sign(A.x-B.x)==0&&sign(A.y-B.y)==0;}
Vector operator - (Point A,Point B){return Vector(A.x-B.x,A.y-B.y);}
double operator ^ (Vector A,Vector B){return A.x*B.y-A.y*B.x;}
double operator * (Vector A,Vector B){return A.x*B.x+A.y*B.y;}
Point D[MAXN],H[MAXN];
int N,L;
int Andrew(int dnt){
int hnt=0,k=0;
sort(D+1,D+dnt+1);
dnt=unique(D+1,D+dnt+1)-D-1;
for(int i=1;i<=dnt;i++){
while(hnt>1&&(sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1])))<=0) hnt--;
H[++hnt]=D[i];
} k=hnt;
for(int i=dnt-1;i>=1;i--){
while(hnt>k&&(sign((H[hnt]-H[hnt-1])^(D[i]-H[hnt-1])))<=0) hnt--;
H[++hnt]=D[i];
}
return hnt;
}
double GL(Vector A){//Get_Length
return sqrt(A*A);
}
double GPC(int hnt){//Get_Polygon_Circumference
double C=0;
for(int i=1;i<hnt;i++) C+=GL(H[i]-H[i+1]);
return C;
}
int main(){
while(~scanf("%d%d",&N,&L)){
for(int i=1;i<=N;i++) D[i].Read();
N=Andrew(N);
printf("%.0f\n",GPC(N)+Pi*2*L);
}
return 0;
}
