[计算几何][求解凸包]Surround the Trees

Problem Description

There are a lot of trees in an area. A peasant wants to buy a rope to surround all these trees. So at first he must know the minimal required length of the rope. However, he does not know how to calculate it. Can you help him?
The diameter and length of the trees are omitted, which means a tree can be seen as a point. The thickness of the rope is also omitted which means a rope can be seen as a line.



There are no more than 100 trees.

 

Input

The input contains one or more data sets. At first line of each input data set is number of trees in this data set, it is followed by series of coordinates of the trees. Each coordinate is a positive integer pair, and each integer is less than 32767. Each pair is separated by blank.

Zero at line for number of trees terminates the input for your program.

 

Output

The minimal length of the rope. The precision should be 10^-2.

 

Sample Input

9 12 7 24 9 30 5 41 9 80 7 50 87 22 9 45 1 50 7 0

 

Sample Output

243.06

 

思路：Graham扫描法求解凸包——https://max.book118.com/html/2017/0409/99594646.shtm(讲得炒鸡生动！！)

AC代码：

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
int n;
int top;
double ans;

struct Point{
  int x,y;
}point[110],stk[110];

double dis(Point a,Point b){//求两点距离
   return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

double cross(Point a,Point b,Point c){//向量ab与向量ac的叉积
   return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}

void find_miny(){//找point[0]
   Point tmp=point[0];
   int flag=0;
   for(int i=1;i<n;i++){
     if(point[i].y<tmp.y||(point[i].y==tmp.y&&point[i].x<tmp.x)){
         tmp=point[i]; flag=i;
     }
   }
   if(flag) swap(point[0],point[flag]);
}

bool cmp(Point a,Point b){//点的排序
   double n=cross(point[0],a,b);
   if(n>0||(n==0&&dis(point[0],a)<dis(point[0],b))) return true;
   return false;
}

void Graham(){//Graham扫描法
   top=-1;
   stk[++top]=point[0]; stk[++top]=point[1];
   for(int i=2;i<n;i++){
      while(top&&cross(stk[top-1],stk[top],point[i])<0) top--;
      stk[++top]=point[i];
   }
}

void sum(){//求凸包的周长
   ans=0;
   stk[++top]=point[0];
   for(int i=0;i<top;i++){
     ans+=dis(stk[i],stk[i+1]);
   }
}

int main()
{
    while(scanf("%d",&n)!=EOF&&n){
        for(int i=0;i<n;i++){
            scanf("%d%d",&point[i].x,&point[i].y);
        }
        if(n==1) {
            printf("0.00
"); continue;
        }
        if(n==2) {
            printf("%.2f
",dis(point[0],point[1])); continue;
        }
        find_miny();
        sort(point+1,point+n,cmp);
        Graham();
        sum();
        printf("%.2f
",ans);
    }
    return 0;
}