hdu 1392(求凸包周长)

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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
   
   
   
   

 

解题思路：求凸包的周长，因为最后栈里面的元素已经是符合要求的凸包上的点，所以直接求这些点的距离和即可。此外要注意n=1和n=2的特殊判断。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;

const int N = 105;
const double eps = 1e-8;
struct Point
{
    double x,y;
} p[N];
int n;
Point Stack[N]; 

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

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

int cmp(Point a,Point b)  ///设<p1,p2,...pm>为对其余点按以p0为中心的极角逆时针排序所得的点集（如果有多个点有相同的极角，除了距p0最远的点外全部移除)
{
    if(mult(a,b,p[0])>0)
        return 1;
    if(mult(a,b,p[0])==0&&dis(b,p[0])-dis(a,p[0])>eps)
        return 1;
    return 0;
}

int Graham()
{
    int top=2; ///栈顶在2，因为凸包的前两个点是不会变了
    sort(p+1,p+n,cmp);
    Stack[0]=p[0];   ///压p0p1p2进栈S
    Stack[1]=p[1];
    Stack[2]=p[2];
    for(int i=3;i<n;i++){
        while(top >= 1 && mult(p[i],Stack[top],Stack[top-1]) >= 0){///有了更好的选择
            top--;
        }
        Stack[++top]=p[i];
    }
    return top;
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n == 0) break;
        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(dis(p[0],p[1])));
            continue;
        }
        int k = 0;
        for(int i=0; i<n; i++) ///令p0为Q中Y-X坐标排序下最小的点
            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]);
        double sum = 0;
        int top = Graham();
        for(int i=1;i<=top;i++){
            sum+=sqrt(dis(Stack[i],Stack[i-1]));
        }
        ///处理最后一个点和P0
        sum+=sqrt(dis(Stack[0],Stack[top]));
		printf("%.2lf\n",sum);
    }
    return 0;
}