链接：传送门

题意：给出二维坐标轴上 n 个点，这 n 个点构成了一个城堡，国王想建一堵墙，城墙与城堡之间的距离总不小于一个数 L ，求城墙的最小长度，答案四舍五入

思路：城墙与城堡直线长度是相等的，当城堡出现拐角时，城墙必然会出现一段圆弧，这些圆弧最终会构成一个半径为 L 的圆，所以答案就是凸包的周长 + 圆的周长

balabala：

1. 采用Jarvis步进法来求凸包，Jarvis步进法复杂度为O(nh)，h为凸包顶点个数
2. 采用Graham-Scan来求凸包，Graham - Scan 法复杂度为O(nlogn)

如有错误请一定指出！

Graham - Scan法：

/*************************************************************************> File Name: hdu1348t2.cpp> Author:    WArobot > Blog:      http://www.cnblogs.com/WArobot/ > Created Time: 2017年05月07日 星期日 20时49分15秒************************************************************************/#include<bits/stdc++.h>
using namespace std;// Graham - Scan
// O(nlgn)#define PI 3.1415926535
const int maxn = 1010;
struct point{double x,y;
};bool cmp(point a,point b){return (a.y<b.y || (a.y==b.y && a.x<b.x));
}
bool mult(point sp,point ep,point op){return (sp.x-op.x)*(ep.y-op.y)>=(sp.y-op.y)*(ep.x-op.x);
}
point res[maxn];
int Graham(point pnt[],int n){int i , len , k = 0 , top = 1;sort(pnt,pnt+n,cmp);if(n == 0)  return 0;   res[0] = pnt[0];if(n == 1)  return 1;   res[1] = pnt[1];if(n == 2)  return 2;   res[2] = pnt[2];for(int i=2;i<n;i++){while( top && mult( pnt[i] , res[top] , res[top-1] ))top--;res[++top] = pnt[i];}len = top;  res[++top] = pnt[n-2];for(i=n-3;i>=0;i--){while( top!=len && mult( pnt[i] , res[top] , res[top-1] ))top--;res[++top] = pnt[i];}return top;
}
double point_dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
int main(){int T , n , L , kase = 0;point pi[maxn];scanf("%d",&T);while(T--){if(kase > 0)    printf("\n");kase++;scanf("%d%d",&n,&L);for(int i=0;i<n;i++)    scanf("%lf%lf",&pi[i].x,&pi[i].y);int t = Graham(pi,n);double ans = 2*PI*L;for(int i=0;i<t;i++){ans += point_dis( res[i] , res[ (i+1)%t ] );}printf("%.lf\n",ans);}return 0;
}

Jarvis步进法：

/*************************************************************************> File Name: hdu1348.cpp> Author:    WArobot > Blog:      http://www.cnblogs.com/WArobot/ > Created Time: 2017年05月07日 星期日 18时55分57秒************************************************************************/#include<bits/stdc++.h>
using namespace std;#define PI 3.1415926535
const int maxn = 1010;
struct point{double x,y;
}pi[maxn];bool cmp(point a,point b){return ( a.y<b.y || a.y==b.y && a.x<b.x);
}int n,L,ans[maxn],cnt,sta[maxn],tail;
// 检查是否严格左转，共线不算左转
bool CrossLeft(point p1,point p2,point p3){ return ((p3.x-p1.x)*(p2.y-p1.y) - (p2.x-p1.x)*(p3.y-p1.y)) < 0;
}
void Jarvis(){tail = cnt = 0; sort(pi,pi+n,cmp);sta[tail++] = 0;    sta[tail++] = 1;for(int i=2;i<n;i++){while(tail>1 && !CrossLeft( pi[ sta[tail-1] ] , pi[ sta[tail-2] ] , pi[i] ))tail--;sta[ tail++ ] = i;}for(int i=0;i<tail;i++) ans[cnt++] = sta[i];tail = 0;   sta[ tail++ ] = n-1;    sta[ tail++ ] = n-2;for(int i=n-3;i>=0;i--){while(tail>1 && !CrossLeft( pi[ sta[tail-1] ] , pi[ sta[tail-2] ] , pi[i] ))tail--;sta[ tail++ ] = i;}for(int i=0;i<tail;i++) ans[cnt++] = sta[i];
}
double Point_dis(point a,point b){return sqrt((a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y));
}
int main(){int T , kase = 0;scanf("%d",&T);while(T--){if(kase)    printf("\n");kase++;scanf("%d%d",&n,&L);for(int i=0;i<n;i++)    scanf("%lf%lf",&pi[i].x,&pi[i].y);Jarvis();double re = 2*PI*L;for(int i=0;i<cnt-1;i++){re += Point_dis( pi[ans[i]] , pi[ans[i+1]] );}printf("%.0lf\n",re);}return 0;
}