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  • 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): 3312 Accepted Submission(s): 1268


    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

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    Ignatius.L
     
    简单的凸包模板题目:
      1 #include<stdio.h>
    2 #include<stdlib.h>
    3 #include<math.h>
    4 typedef struct
    5 {
    6 double x , y ;
    7 }POINT ;
    8 POINT result[110] ;// 模拟堆栈S,保存凸包上的点
    9 POINT tree[110] ;
    10 int n , top ;
    11 double Distance ( POINT p1 , POINT p2 )
    12 {
    13 return sqrt( (p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y) ) ;
    14 }
    15 double Multiply(POINT p1 , POINT p2 , POINT p3) // 叉积
    16 {
    17 return ( (p2.x - p1.x)*(p3.y - p1.y) - (p2.y - p1.y)*(p3.x - p1.x) ) ;
    18 }
    19 int cmp ( const void *p1 , const void *p2 )
    20 {
    21 POINT *p3,*p4;
    22 double m;
    23 p3 = (POINT *)p1;
    24 p4 = (POINT *)p2;
    25 m = Multiply(tree[0] , *p3 , *p4) ;
    26 if(m < 0) return 1;
    27 else if(m == 0 && (Distance(tree[0] , *p3) < Distance(tree[0],*p4)))
    28 return 1;
    29 else return -1;
    30 }
    31 void Tubao ()
    32 {
    33 int i ;
    34 result[0].x = tree[0].x;
    35 result[0].y = tree[0].y;
    36 result[1].x = tree[1].x;
    37 result[1].y = tree[1].y;
    38 result[2].x = tree[2].x;
    39 result[2].y = tree[2].y;
    40 top = 2;
    41 for ( i = 3 ; i <= n ; ++ i )
    42 {
    43 while (Multiply(result[top - 1] , result[top] , tree[i]) <= 0 )
    44 top -- ; //出栈
    45 result[top + 1].x = tree[i].x ;
    46 result[top + 1].y = tree[i].y ;
    47 top ++ ;
    48 }
    49
    50 }
    51 int main ()
    52 {
    53 int pos ;
    54 double len , temp , px , py ;
    55 while ( scanf ( "%d" , &n ) != EOF , n )
    56 {
    57 py = -1 ;
    58 for ( int i = 0 ; i < n ; ++ i )
    59 {
    60 scanf ( "%lf%lf" , &tree[i].x , &tree[i].y ) ;
    61
    62 }
    63 if ( n == 1 )
    64 {
    65 printf ( "0.00\n" ) ;
    66 continue ;
    67 }
    68 else if ( n == 2 )
    69 {
    70 printf ( "%.2lf\n" , Distance(tree[0] , tree[1]) ) ;
    71 continue ;
    72 }
    73 for ( int i = 0 ; i < n ; ++ i )
    74 {
    75 if(py == -1 || tree[i].y < py)
    76 {
    77 px = tree[i].x;
    78 py = tree[i].y;
    79 pos = i;
    80 }
    81 else if(tree[i].y == py && tree[i].x < px)
    82 {
    83 px = tree[i].x;
    84 py = tree[i].y;
    85 pos = i;
    86 }
    87 }
    88 temp = tree[0].x ; // 找出y最小的点
    89 tree[0].x = tree[pos].x ;
    90 tree[pos].x = temp ;
    91 temp = tree[0].y ;
    92 tree[0].y = tree[pos].y ;
    93 tree[pos].y = temp ;
    94 qsort(&tree[1],n - 1,sizeof(double) * 2,cmp);
    95 tree[n].x = tree[0].x;
    96 tree[n].y = tree[0].y;
    97 Tubao();
    98 len = 0.0;
    99 for(int i = 0 ; i < top ; i ++)
    100 len = len + Distance(result[i] , result[i+1]) ;
    101 printf("%.2lf\n",len);
    102
    103 }
    104 return 0 ;
    105 }

      

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  • 原文地址:https://www.cnblogs.com/jbelial/p/2128624.html
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