Surround the Trees-优快云博客

本文链接：https://blog.youkuaiyun.com/czhprogram/article/details/10408451

本文介绍了一种计算包围一组树木所需的最短绳长的方法。通过将树木视为平面上的点，利用计算几何中的凸包算法确定了绳子的最小长度。文章提供了完整的C++代码实现，并附带了一个示例输入和输出。

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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

Sample Output

243.06

求凸包

#include<iostream>
#include <iomanip>
#include<algorithm>
#include<cmath>
using namespace std;
struct Point{
int x;
int y;
}p[1000000];
Point queue[10000];
Point start;
double distance1(Point a,Point b)
{

return sqrt(pow(b.x-a.x,2)+pow(b.y-a.y,2));

}
double direction(Point a,Point b,Point c)
{
return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);

}

bool cmp(const Point a,const Point b)
{

int k=(a.x-start.x)*(b.y-start.y)-(b.x-start.x)*(a.y-start.y);
if(k>0)
{
return true;

}
if(k==0)
{
if(distance1(start,a)<distance1(start,b))
return true;

}
return false;

}
void ConvexB(int n)
{
queue[0]=p[0];
queue[1]=p[1];
queue[2]=p[2];
int count=3;
int top=2;
for(int i=3;i<n;i++)
{

while(count>=2)
{
Point nexttop=queue[top-1];
Point front=queue[top];
int dir=direction(nexttop,front,p[i]);
if(dir>0)
{
break;
}
top--;
count--;

}
top++;
count++;
queue[top]=p[i];

}
if(top>=2)
{
double re=0.0;
for(int j=0;j<top;j++)
{
re+=distance1(queue[j],queue[j+1]);

}
re+=distance1(queue[top],queue[0]);
printf("%.2lf\n",re);
}

}

int main()
{
//freopen("in.txt","r",stdin);

int n;
while(scanf("%d",&n)!=EOF)
{
if(n==0)
break;
int i=1;
int px=-1;
int py=-1;
int p1;
for(i=0;i<n;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
if(py==-1||p[i].y<py)
{
px=p[i].x;
py=p[i].y;
p1=i;


}
if(p[i].y==py)
{
if(p[i].x<px)
{

px=p[i].x;
py=p[i].y;
p1=i;

}

}



}
if(n<=1)
{
cout<<0.00<<endl;
continue;
}
if(n==2)
{
printf("%.2lf\n",distance1(p[1],p[0]));
continue;

}
Point temp=p[0];
p[0]=p[p1];
p[p1]=temp;

start=p[0];
sort(p+1,p+n,cmp);
ConvexB(n);

}
return 0;

}