acm几何凸包hdu1392

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

这是一个凸包模板的基础题，刚开始在continue和break错了好长时间，下面是正确代码。

#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
struct point
{
    double x,y;
}pnt[10001],res[10001];
// 两向量叉积
double cross( point sp,point ep,point op)
{
    return (sp.x-op.x)*(ep.y-op.y)-(ep.x-op.x)*(sp.y-op.y);
}
// sort数组排序，比较的<号重载
bool operator < (const point &l,const point &r)
{
    return l.y<r.y || (l.y==r.y && l.x<r.x) ;
}
// 求两点间几何距离
double dis(point a,point b)
{
    return sqrt( (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y) );
}
int Graham( int n )
{
    int i,len,top=1;
    // 先对所有点按照极角排序
    // 纵坐标最小的排在前面，纵坐标相同，横坐标小的排在前面
    sort(pnt,pnt+n);
    // 判断点的个数是否大于2（所给的点能否构成凸包）
    if( n==0 )  return 0;res[0] = pnt[0];
    if( n==1 )  return 1;res[1] = pnt[1];
    if( n==2 )  return 2;res[2] = pnt[2];
    // 用叉积来判断后面的点是否为凸包中的点
    for( i=2;i<n;++i )
    {
        while( top && cross(pnt[i],res[top],res[top-1])>=0 )   top--;
        res[++top] = pnt[i];
    }
    len = top; res[++top] = pnt[n-2];
    // 判断最后三个点
    for(i=n-3;i>=0;--i)
    {
        while( top!=len && cross(pnt[i],res[top],res[top-1])>=0 )  top--;
        res[++top] = pnt[i];
    }
    return top;
}
int main()
{
    int n,len;
    double sum;
    while(scanf("%d",&n)!=-1)
    {
        for(int i=0;i<n;i++)
        scanf("%lf%lf",&pnt[i].x,&pnt[i].y);
        if(n==0)
        break;
        len=Graham(n);
        if(len==0||len==1)
        {printf("0\n");continue;}
        if(len==2)
        {printf("%.2lf\n",dis(res[0],res[1]));continue;}
        sum=0;
        for(int i=0;i<len;i++)
        sum+=dis(res[i],res[i+1]);
        printf("%.2lf\n",sum);


    }
}