Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10345    Accepted Submission(s): 4009


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
 

 

Source
Asia 1997, Shanghai (Mainland China)
 
题意：n个点 求凸包的周长
题解：凸包模板题
  1 /******************************
  2 code by drizzle
  3 blog: www.cnblogs.com/hsd-/
  4 ^ ^    ^ ^
  5  O      O
  6 ******************************/
  7 #include<bits/stdc++.h>
  8 #include<map>
  9 #include<set>
 10 #include<cmath>
 11 #include<queue>
 12 #include<bitset>
 13 #include<math.h>
 14 #include<vector>
 15 #include<string>
 16 #include<stdio.h>
 17 #include<cstring>
 18 #include<iostream>
 19 #include<algorithm>
 20 #pragma comment(linker, "/STACK:102400000,102400000")
 21 using namespace std;
 22 #define  A first
 23 #define B second
 24 const int mod=1000000007;
 25 const int MOD1=1000000007;
 26 const int MOD2=1000000009;
 27 const double EPS=0.00000001;
 28 typedef __int64 ll;
 29 const ll MOD=1000000007;
 30 const int INF=1000000010;
 31 const ll MAX=1ll<<55;
 32 const double eps=1e-8;
 33 const double inf=~0u>>1;
 34 const double pi=acos(-1.0);
 35 typedef double db;
 36 typedef unsigned int uint;
 37 typedef unsigned long long ull;
 38 struct point
 39 {
 40     double x,y;
 41     point(double x=0,double y=0):x(x),y(y) {}
 42 };
 43 typedef point vec;
 44 vec operator -(point a,point b)
 45 {
 46     return vec(a.x-b.x,a.y-b.y);
 47 }
 48 vec operator +(point a,point b)
 49 {
 50     return vec(a.x+b.x,a.y+b.y);
 51 }
 52 vec operator *(point a,double t)
 53 {
 54     return vec(a.x*t,a.y*t);
 55 }
 56 vec operator /(point a,double t)
 57 {
 58     return vec(a.x/t,a.y/t);
 59 }
 60 int dcmp(double x)
 61 {
 62     if(fabs(x)<=eps) return 0;
 63     return x<0?-1:1;
 64 }
 65 double cross(vec a,vec b)  ///叉积
 66 {
 67     return a.x*b.y-a.y*b.x;
 68 }
 69 bool cmp(point a,point b)
 70 {
 71     if(fabs(a.x-b.x)<=eps) return a.y<b.y;
 72     return a.x<b.x;
 73 }
 74 double disn(point a,point b)
 75 {
 76     return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
 77     /*两点之间的距离*/
 78 }
 79 void convexhull(point *s,int &n)
 80 {
 81     sort(s,s+n,cmp);
 82     int m=0;
 83     point p[110];
 84     for(int i=0; i<n; i++)
 85     {
 86         while(m>1 && dcmp(cross(p[m-1]-p[m-2],s[i]-p[m-2]))<=0)
 87             m--;
 88         p[m++]=s[i];
 89     }
 90     int k=m;
 91     for(int i=n-2; i>=0; i--)
 92     {
 93         while(m>k && dcmp(cross(p[m-1]-p[m-2],s[i]-p[m-2]))<=0)
 94             m--;
 95         p[m++]=s[i];
 96     }
 97     m--;
 98     n=m;
 99     for(int i=0; i<n; i++) s[i]=p[i];
100     /*建立凸包*/
101 }
102 int jishu;
103 int main()
104 {
105     while(scanf("%d",&jishu)!=EOF)
106     {
107         if(jishu==0)
108             break;
109         point s[110];
110         for(int i=0; i<jishu; i++)
111         {
112             scanf("%lf %lf",&s[i].x,&s[i].y);
113 
114         }
115         if(jishu==1)
116         {
117             printf("0.00\n");
118                 continue;
119         }
120         if(jishu==2)
121         {
122             printf("%.2f\n",disn(s[0],s[1]));
123                continue;
124         }
125         convexhull(s,jishu);
126         double sum=0;
127         for(int i=0; i<jishu-1; i++)
128         {
129             sum+=disn(s[i],s[i+1]);
130         }
131         sum+=disn(s[jishu-1],s[0]);
132         printf("%.2f\n",sum);
133     }
134     return 0;
135 }