Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10403    Accepted Submission(s): 4033

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：

求以上n个点的凸包的周长

讲解很详细的博客：http://www.cnblogs.com/jbelial/archive/2011/08/05/2128625.html

代码：

  1 //求凸包的模板题Graham扫描法。
  2 //详解《算法导论》604页
  3 //极角排序先比较象限再比较叉积。
  4 #include<iostream>
  5 #include<cstdio>
  6 #include<cstring>
  7 #include<algorithm>
  8 #include<cmath>
  9 using namespace std;
 10 int top,n;
 11 struct nod
 12 {
 13     double x,y;
 14 } p[102],que[102];  //que用于保存组成凸包的点
 15 int xy(nod p0)   //返回点相对于p[0]点所在的象限
 16 {
 17     p0.x-=p[0].x;
 18     p0.y-=p[0].y;
 19     if(p0.x>=0&&p0.y>=0) return 1;
 20     if(p0.x<=0&&p0.y>0) return 2;
 21     if(p0.x<0&&p0.y<=0) return 3;
 22     if(p0.x>=0&&p0.y<0) return 4;
 23 }
 24 double dis(nod p1,nod p2)    //计算凸包边长
 25 {
 26     return sqrt((p2.x-p1.x)*(p2.x-p1.x)+(p2.y-p1.y)*(p2.y-p1.y));
 27 }
 28 double chaji(nod p0,nod p1,nod p2)  //叉积
 29 {
 30     return ((p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x));
 31 }
 32 bool cmp(nod p1,nod p2)
 33 {
 34     int L1=xy(p1),L2=xy(p2);
 35     if(L1==L2)
 36     {
 37         int tem=chaji(p[0],p1,p2);
 38         if(tem>0) return 1;   //tem>0说明向量p1p0在向量p2p0的顺时针方向即p1p0相对于p0的极角小于p2p0的
 39         if(tem<0) return 0;
 40         if(tem==0)
 41             return p1.x<p2.x;
 42     }
 43     else return L1<L2;
 44 }
 45 void tubao()
 46 {
 47     top=0;
 48     que[top].x=p[0].x;
 49     que[top++].y=p[0].y;
 50     que[top].x=p[1].x;
 51     que[top++].y=p[1].y;
 52     que[top].x=p[2].x;
 53     que[top].y=p[2].y;
 54     for(int i=3; i<=n; i++)
 55     {
 56         while(chaji(que[top-1],que[top],p[i])<=0)
 57             top--;
 58         que[++top].x=p[i].x;
 59         que[top].y=p[i].y;
 60     }
 61 }
 62 int main()
 63 {
 64     while(scanf("%d",&n)&&n)
 65     {
 66         int Miny=10000007,Minx=10000007,Mini;
 67         for(int i=0; i<n; i++)
 68         {
 69             scanf("%lf%lf",&p[i].x,&p[i].y);
 70             if(p[i].y<Miny)
 71             {
 72                 Miny=p[i].y;
 73                 Mini=i;
 74             }
 75             else if(p[i].y==Miny&&p[i].x<Minx)
 76             {
 77                 Minx=p[i].x;
 78                 Mini=i;
 79             }
 80         }
 81         if(n==1) {printf("0.00\n");continue;}
 82         if(n==2) {printf("%.2lf\n",dis(p[0],p[1]));continue;}  //计算凸包的点数必须多于2
 83         int tem=p[0].x;
 84         p[0].x=p[Mini].x;
 85         p[Mini].x=tem;
 86         tem=p[0].y;
 87         p[0].y=p[Mini].y;
 88         p[Mini].y=tem;
 89         sort(p+1,p+n,cmp);
 90         p[n].x=p[0].x;p[n].y=p[0].y; //形成闭合的凸包
 91         tubao();
 92         double ans=0.0;
 93         for(int i=0;i<top;i++)
 94         {
 95             ans+=dis(que[i],que[i+1]);
 96         }
 97         printf("%.2lf\n",ans);
 98     }
 99     return 0;
100 }