7.1.4 Surround the Trees

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Surround the Trees

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

Total Submission(s): 166 Accepted Submission(s): 81

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：求凸包

当点数=1、2、3时分类讨论

其余的求凸包，我用的是极角排序（按叉积排），跑的慢啊。。

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#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
using namespace std;

typedef double dd;
const int maxn=110;
const dd eps=1E-8;
struct point
{
    dd x,y;
};
point p[maxn],st[maxn];
int n,cnt;
dd t,ans,x[maxn],y[maxn],mx,my;


void close()
{
exit(0);
}

int dblcmp(dd t)
{
    if (fabs(t)<eps)
        return 0;
    if (t<0) return -1;
    else    return 1;
}

dd dist(point a,point b)
{
    x[1]=fabs(a.x-b.x);
    y[1]=fabs(a.y-b.y);
    return (x[1]*x[1]+y[1]*y[1]);
}

dd mul(point a,point b,point c)
{
    x[1]=b.x-a.x; y[1]=b.y-a.y;
    x[2]=c.x-a.x; y[2]=c.y-a.y;
    return (x[1]*y[2]-y[1]*x[2]);
}

bool cmp(point a,point b)
{
    dd t=mul(st[1],a,b);
    int j=dblcmp(t);
    if (j==0) //共线
        return dist(a,st[1])<dist(b,st[1]);
    if (j==1)
        return true;
    else
        return false;
    return false;
}

bool judge(point a,point b,point c)
{
    dd t=mul(c,b,a);
    int j=dblcmp(t);
    //j==1 顺时针
    if (j==0) return true; //gong xian de ye bu yao
    if (j==1) return false;
    else    return true;
}
void work()
{
    ans=0;
    if (n==1) {ans=0;return;}
    if (n==2) {ans=sqrt(dist(p[1],p[2]));return;}
    if (n==3) {ans=sqrt(dist(p[1],p[2])) + sqrt(dist(p[2],p[3])) + sqrt(dist(p[1],p[3]));return;}
    mx=p[1].x;
    my=p[1].y;
    for (int i=1;i<=n;i++)
    {
        if ( (my==p[i].y && mx>p[i].x) || my>p[i].y)
        {
            mx=p[i].x;
            my=p[i].y;
        }
    }
    st[1].x=mx;
    st[1].y=my;
    sort(p+1,p+n+1,cmp);
    st[2].x=p[2].x;
    st[2].y=p[2].y;
    cnt=2;
    for (int i=3;i<=n;i++)
    {
        while (judge(p[i],st[cnt],st[cnt-1]))
              cnt--;
        cnt++;
        st[cnt].x=p[i].x;
        st[cnt].y=p[i].y;
    }
    for (int i=2;i<=cnt;i++)
        ans+=sqrt(dist(st[i],st[i-1]));
    ans+=sqrt(dist(st[cnt],st[1]));
}

void init()
{
    while (scanf("%d",&n)!=EOF)
    {
        if (n==0) close();
        for (int i=1;i<=n;i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }
        work();
        printf("%.2lf
",ans);
    }
}

int main ()
{
    init();
    close();
    return 0;
}

http://blog.chinaunix.net/uid-22263887-id-1778926.html

http://www.cnblogs.com/Booble/archive/2011/03/10/1980089.html#2065991

http://blog.csdn.net/javasui/article/details/6759777

http://blog.sina.com.cn/s/blog_87cb8e680100svnd.html

http://blog.chinaunix.net/uid-22263887-id-1778926.html

http://www.cnblogs.com/devymex/archive/2010/08/09/1795392.html