hdu-1392 Surround the Trees(凸包)

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Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9720    Accepted Submission(s): 3730

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
   
   
   
   

题意：有很多棵树，求一根最短的绳子，绑在某些树上，能使所有树都在绳子内。

思路：明显的凸包模板，只是这道题十分坑。n==1和n==2要特判，因为一般凸包题都是从3开始的。但这道题没有说明，而且也确实有n==2的数据。

注意用的是绳子n==2的时候不需要乘以二，只要在两颗树上绑一根绳子便好。已经能满足两颗树都在绳子内的要求。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define eps 1e-8
#define N 110
struct Node
{
    double x,y;
}p[N],stack[N];
double mulit(Node a,Node b,Node c)
{
    return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y);
}
double dist(Node a,Node b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
int cmp(Node a,Node b)
{
    if(mulit(p[0],a,b)>0)
        return 1;
    if(mulit(p[0],b,a)==0&&(dist(p[0],b)-dist(p[0],a))<eps)
        return 1;
    return 0;
}
int Graham(int n)
{
    sort(p+1,p+n,cmp);
    stack[0]=p[0];
    stack[1]=p[1];
    stack[2]=p[2];
    int top=2;
    for(int i=3;i<n;i++)
    {
        while(top>=1&&mulit(stack[top-1],p[i],stack[top])>=0)
            top--;
            stack[++top]=p[i];
    }
    return top;
}
int main()
{
    int n;
    while(~scanf("%d",&n)&&n)
    {
        for(int i=0;i<n;i++)
            scanf("%lf %lf",&p[i].x,&p[i].y);
        if(n==1)
        {
            printf("0.00\n");
            continue;
        }
        if(n==2)
        {
            printf("%.2lf\n",sqrt(dist(p[0],p[1])));
            continue;
        }
        int k=0;
        for(int i=0;i<n;i++)
        {
            if(p[k].y>p[i].y||(p[k].y==p[i].y&&p[k].x>p[i].x))
                k=i;
        }
        swap(p[0],p[k]);
        int top=Graham(n);
        double sum=0.0;
        for(int i=1;i<=top;i++)
            sum+=sqrt(dist(stack[i],stack[i-1]));
            sum+=sqrt(dist(stack[0],stack[top]));
            printf("%.2lf\n",sum);
    }
    return 0;
}