HDU 1392 Surround the Trees .

Surround the Trees

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3223    Accepted Submission(s): 1230

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)

 

 

Recommend

Ignatius.L

 

凸包水题直接GramhamScan可以过，不过要注意n < 3两个特例。

果断贴代码：

4285845

2011-07-29 12:51:30

Accepted

1392

78MS

272K

1631 B

C++

10SGetEternal{（。）（。）}!

#include <iostream>
#include <vector>
#include <cmath>
#include <cstdlib>
#include <algorithm>
using namespace std;
#define MAXI 111

class point
{
public :
    double x, y;

    point(double tx = 0.0, double ty = 0.0)
    {
        x = tx;
        y = ty;
    }

    double dis() 
    { 
        return sqrt(pow(x, 2) + pow(y, 2)); 
    };

    point operator - (point &o)
    {
        return point(x - o.x, y - o.y);
    }

    double operator * (point &o) 
    {
        return x * o.y - y * o.x;
    }
}ps[MAXI];

vector<point> chul;
int n;

bool cmp(point &a, point &b)
{
    double cp = (ps[0] - a) * (b - a);

    if (cp == 0) return (a - ps[0]).dis() > (b - ps[0]).dis();
    else  if (cp < 0)  return 1;
    return 0;
}

double GHS()
{
    int i, idm;
    double tsu = 0.0;

    for (idm = i = 0; i < n; i++)
        if (ps[i].y == ps[idm].y && ps[i].x < ps[idm].x || ps[i].y < ps[idm].y)
            idm = i;
    swap(ps[idm], ps[0]);
    sort(ps + 1, ps + n, cmp);
    chul.clear();
    chul.push_back(ps[0]);
    chul.push_back(ps[1]);
    for (i = 2; (ps[0] - ps[1]) * (ps[i] - ps[1]) == 0; i++);
    chul.push_back(ps[i]);
    for (i++; i < n; i++)
    {
        while ((chul[chul.size() - 2] - chul[chul.size() - 1]) * (ps[i] - chul[chul.size() - 1]) >= 0)
            chul.pop_back();
        chul.push_back(ps[i]);
    }
    for (i = 0; i < chul.size(); i++)
        tsu += (chul[(i + 1) % chul.size()] - chul[i]).dis(); 

    return tsu;
}

int main()
{
    int i;

    while (scanf("%d", &n), n)
    {
        for (i = 0; i < n; i++)
            scanf("%lf%lf", &ps[i].x, &ps[i].y);
        if (n == 2) printf("%.2lf\n", ps[1] - ps[0]);
        else if (n ==  1) printf("0\n");
        else printf("%.2lf\n", GHS());
    }

    return 0;
}

 Thanks……