Problem--HOJ1929 Hardwood Species(Binary Search Tree)

Problem--HOJ1929 Hardwood Species(Binary Search Tree)

Analysis:
Given a list of species of trees,we are required to print out a list containing the names of species together with the percentage of each,within the whole.
The idea of this is quite simple.Input all the species and put them in a set,while counting how many times each element appears.However,to reach this goal within certain time is not that easy a task.And this is when the data structure Binary Search Tree functions well.We can simply maintain a Binary Search Tree here,and output the result when the insert process is finished.See the code below for details.
Code:

 1  #include < cstdio >
 2  #include < string >
 3  #include < iostream >
 4  #include < cstring >
 5  using   namespace  std;
 6  double  sum;
 7  struct  tree
 8  {
 9       char  name[ 51 ];
10       double  num;
11      tree *  lchild;
12      tree *  rchild;
13  }tmp, * Root,pool[ 10000 ], * pp;
14  tree *  Insert_AVL(tree *  Root, char   * str)
15  {
16       if (Root == NULL)
17      {
18          strcpy(pp -> name,str);pp -> num = 1 ;
19          pp -> lchild = NULL;pp -> rchild = NULL;
20          Root = pp;pp ++ ;
21      }
22       else
23      {
24           int  judge = strcmp(str,Root -> name);
25           if (judge == 0 ) Root -> num ++ ;
26           else   if (judge < 0 )
27              Root -> lchild = Insert_AVL(Root -> lchild,str);
28           else
29              Root -> rchild = Insert_AVL(Root -> rchild,str);
30      }
31       return  Root;
32  }
33  void  Inorder(tree *  Root)
34  {
35       if (Root == NULL)  return ;
36       if (Root -> lchild) Inorder(Root -> lchild);
37       if (Root)
38          printf( " %s %.4lf\n " ,Root -> name,((Root -> num) / sum) * 100.0 );
39       if (Root -> rchild) Inorder(Root -> rchild);
40  }
41  int  main()
42  {
43      Root = NULL;pp = pool;sum = 0 ;
44       char  str[ 51 ];
45       while (gets(str) != NULL)
46      {
47           if (str[ 0 ] != ' \0 ' )
48          {
49              Root = Insert_AVL(Root,str);sum ++ ;
50          }
51           else
52          {
53              Inorder(Root);putchar( ' \n ' );
54              sum = 0 ;pp = pool;Root = NULL;
55          }
56      }
57      Inorder(Root);
58       return   0 ;
59  }