HDUJ 1392 Surround the Trees 凸包

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7203    Accepted Submission(s): 2752

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

水平序的Andrew算法：

#include<iostream>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node{double x,y;}a[105],b[105];double cmp(node n,node m)    //先比较X坐标，在比较Y坐标（从小到大）{if(n.x != m.x)   return  n.x < m.x;elsereturn  n.y < m.y;}double Cross(node a,node b,node c)     //计算叉积大小{return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);}double dis(node a,node b)     //计算距离{return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int CH(node* a,int n,node* b){sort(a,a+n,cmp);int m=0,i;for(i=0;i<n;i++)    //从左往右，选下边界{while(m > 1 && Cross(b[m-2],b[m-1],a[i]) < 0)m--;b[m++]=a[i];}int k=m;for(i=n-2;i>=0;i--)    //从右往左，选上边界{while(m > k && Cross(b[m-2],b[m-1],a[i]) < 0)m--;b[m++]=a[i];}if(n >1)  m--;return m;}int main(){int n;while(cin>>n){if(n==0)   break;memset(a,0,sizeof(a));memset(b,0,sizeof(b));int i,j;for(i=0;i<n;i++){cin>>a[i].x>>a[i].y;}//cout<<CH(a,n,b)<<endl;     //输出所选点的总数if(n==1)   cout<<0.00<<endl;else  if(n==2)printf("%.2lf\n",dis(a[0],a[1]));else{int m=CH(a,n,b);double s=0;for(i=1;i<m;i++)s+=dis(b[i-1],b[i]);s+=dis(b[0],b[m-1]);printf("%.2lf\n",s);}//for(i=0;i<CH(a,n,b);i++)    //输出所选点的坐标//cout<<b[i].x<<" "<<b[i].y<<endl;   }return 0;}