UVa 11234 Expressions

UVa 11234 Expressions

题目要求重新写出给出的后缀表达式，使原后缀表达式的栈方法和新表达式的队列方法表示的表达式相同，也就是两种表达式对应的表达式树相等。
仔细想想表达式树和队列的结构，可以发现，输出为表达式树按层次遍历得到的序列的逆序。
以下是我的代码：

#include < stack >
#include < queue >
#include < cstdio >
#include < cstring >
using   namespace  std;
const   int  kMaxl( 10007 );

struct  Node
{
     int  father,left,right;
};

char  r[kMaxl];
Node tree[kMaxl];
int  ans[kMaxl];

int  main()
{
     /*
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    // */

     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
        scanf( " %s " ,r);

         int  n(strlen(r));
        stack < int >  s;
        memset(tree, - 1 ,kMaxl * sizeof (Node));
         for ( int  i = 0 ;i < n;i ++ )
        {
             if (r[i] >= ' a '   &&  r[i] <= ' z ' )
                s.push(i);
             else
            {
                 int  x,y;
                y = s.top();
                s.pop();
                x = s.top();
                s.pop();
                tree[i].left = x;
                tree[i].right = y;
                tree[x].father = tree[y].father = i;
                s.push(i);
            }
        }

         /*
        for(int i=0;i<n;i++)
            printf("%d %d %d\n",tree[i].father,tree[i].left,tree[i].right);
        // */

         int  root;
         for ( int  i = 0 ;i < n;i ++ )
             if (tree[i].father ==- 1 )
            {
                root = i;
                 break ;
            }
        queue < int >  q;
         int  len = 0 ;
        memset(ans, 0 ,kMaxl * sizeof ( int ));
        q.push(root);
         while ( ! q.empty())
        {
             int  t(q.front());
            q.pop();
            ans[len ++ ] = t;
             if (tree[t].left !=- 1 )
                q.push(tree[t].left);
             if (tree[t].right !=- 1 )
                q.push(tree[t].right);
        }
         for ( int  i = len - 1 ;i >= 0 ;i -- )
            printf( " %c " ,r[ans[i]]);
        printf( " \n " );
    }

     return   0 ;
}