hdu 1392 Surround the Trees(求取凸包并求凸包的周長)

Surround the Trees

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

Total Submission(s): 8414    Accepted Submission(s): 3222

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) 題目分析：

因為要将所有樹都圍起來繩的最短長度，可以假設出現凹包，情況，那麼将凹角出現的點去掉，直接相連另兩個點，因為三角形兩邊之和大于第三邊，是以圍成凸包的周長就是最小的繩的長度

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define MAX 107

using namespace std;

int n;

struct Node
{
    int x,y;
}e[MAX],res[MAX];

int cmp ( Node a , Node b )
{
    if ( a.x == b.x ) return a.y < b.y;
    return a.x < b.x;
}

int cross ( Node a , Node b , Node c )
{
    return (a.x-c.x)*(b.y-c.y) - (b.x-c.x)*(a.y-c.y);
}

int convex ( int n )
{
    sort ( e , e+n , cmp );
    int m = 0 , i , j , k;
    for ( int i = 0 ; i < n ; i++ )
    {
        while ( m > 1 && cross ( res[m-1] , e[i] , res[m-2] ) <=0 ) m--;
        res[m++] = e[i];
    }
    k = m;
    for ( int i = n-2 ; i >= 0 ; i-- )
    {
        while ( m > k && cross ( res[m-1] , e[i] , res[m-2] ) <= 0 ) m--;
        res[m++] = e[i];
    }
    if ( n > 1 ) m--;
    return m;
}

double dis ( Node p1 , Node p2 )
{
    return sqrt (pow ( (p1.x - p2.x)*1.0 , 2 ) + pow ( (p1.y-p2.y)*1.0 , 2 ) );
}

double perimeter ( int n )
{
    double ret = 0;
    if ( n < 2 ) return ret;
    if ( n == 2 ) return dis ( res[0] , res[1] );
    Node p1 = res[0],p2,p3 = res[0];
    for ( int i = 1 ; i < n ; i++ )
    {
        p2 = res[i];
        ret += dis ( p1 , p2 );
        p1 = p2;  
    }
    ret += dis ( p1 , p3 );
    return ret;
}

int main ( )
{
    while ( ~scanf ( "%d" , &n ) , n )
    {
        for ( int i = 0 ; i < n ; i++ )
            scanf ( "%d%d" , &e[i].x , &e[i].y );
        printf ( "%.2lf\n" , perimeter ( convex ( n ) ) );
    } 
}