题目地址：POJ 1113

先求出凸包的周长，然后剩下的弧合起来一定是个半径为l的圆，然后再加上以l为半径的圆的周长即可。

代码如下：

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdlib.h>
#include <math.h>
#include <ctype.h>
#include <queue>
#include <map>
#include <set>
#include <algorithm>

using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
int n, top, l;
double PI=acos(-1.0);
struct Point
{
    int x, y;
}p[1010], tu[1010];
double dist(Point x, Point y)
{
    return sqrt((x.x-y.x)*(x.x-y.x)*1.0+(x.y-y.y)*(x.y-y.y));
}
Point operator - (Point x, Point y)
{
    Point z;
    z.x=x.x-y.x;
    z.y=x.y-y.y;
    return z;
}
int Cross(Point x, Point y)
{
    return x.x*y.y-x.y*y.x;
}
int cmp(Point x, Point y)
{
    if(x.x==y.x) return x.y<y.y;
    return x.x<y.x;
}
void Andew()
{
    int i, j, k;
    sort(p,p+n,cmp);
    top=0;
    for(i=0;i<n;i++)
    {
        while(top>1&&Cross(tu[top-1]-tu[top-2],p[i]-tu[top-1])<0) top--;
        tu[top++]=p[i];
    }
    k=top;
    for(i=n-2;i>=0;i--)
    {
        while(top>k&&Cross(tu[top-1]-tu[top-2],p[i]-tu[top-1])<0) top--;
        tu[top++]=p[i];
    }
    double ans=0;
    for(i=1;i<top;i++)
    {
        ans+=dist(tu[i],tu[i-1]);
        //printf("%.2f\n",ans);
    }
    ans+=2*PI*l;
    printf("%d\n",(int)(ans+0.5));
}
int main()
{
    int i, j;
    scanf("%d%d",&n,&l);
    for(i=0;i<n;i++)
    {
        scanf("%d%d",&p[i].x,&p[i].y);
    }
    Andew();
    return 0;
}

POJ 1113 Wall (凸包)