凸包(hd1392)

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8996    Accepted Submission(s): 3457
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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
9
12 7
24 9
30 5
41 9
80 7
50 87
22 9
45 1
50 7
0
 
 
Sample Output
243.06
 1 #include <iostream>
 2 #include <algorithm>
 3 #include <cmath>
 4 #include <cstdio>
 5 #include <cstring>
 6 using namespace std;
 7 struct point
 8 {
 9     int x,y;
10 }list[100];
11 int stack[100];
12 int cross(point p0,point p1,point p2)
13 {
14     return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x); 
15 }
16 double dis(point p1,point p2)
17 {
18     return sqrt(double(p2.x-p1.x)*(p2.x-p1.x)+double(p2.y-p1.y)*(p2.y-p1.y));
19 }
20 bool cmp(point p1,point p2)
21 {
22     int tem=cross(list[0],p1,p2);
23     if(tem>0)
24         return 1;
25     else if(tem==0&&dis(list[0],p1)<dis(list[0],p2))
26         return 1;
27     else
28         return 0;
29 }
30 void init(int n)
31 {
32     int i,k;
33     point p0;
34     cin>>list[0].x>>list[0].y;
35     for(i=1;i<n;i++)
36     {
37         cin>>list[i].x>>list[i].y;
38         if(list[i].y<list[0].y||(list[i].y==list[0].y&&list[i].x<list[0].x))
39         {
40             p0=list[0];
41             list[0]=list[i];
42             list[i]=p0;
43         }
44     }
45     sort(list+1,list+n,cmp);
46 }
47 int main()
48 {
49     int i,n;
50     //freopen("in.txt","r",stdin);
51     while(cin>>n&&n)
52     {
53         memset(list,0,sizeof(list));
54         memset(stack,0,sizeof(stack));
55         if(n==1)
56             cout<<0.00<<endl;
57         else if(n==2)
58         {
59             cin>>list[0].x>>list[0].y>>list[1].x>>list[1].y;
60             printf("%0.2f\n",dis(list[0],list[1]));
61         }
62         else
63         {
64             stack[0]=0;
65             stack[1]=1;
66             int top=1;
67             init(n);
68             for(i=2;i<n;i++)
69             {
70                 while(top>0&&cross(list[stack[top]],list[stack[top-1]],list[i])>0)
71                     top--;
72                 top++;
73                 stack[top]=i;
74             }
75             double sum=0;
76             for(i=1;i<=top;i++)
77             {
78                 sum+=dis(list[stack[i]],list[stack[i-1]]);
79             }
80             sum+=dis(list[0],list[stack[top]]);
81             printf("%0.2f\n",sum);
82             }
83         }
84 }

 

 
posted @ 2015-09-16 21:23  御心飞行  阅读(154)  评论(0)  编辑  收藏  举报