Anagrams by Stack

Description

How can anagrams result from sequences of stack operations? There are two sequences of stack operators which can convert TROT to TORT:

[
i i i i o o o o
i o i i o o i o
]

where i stands for Push and o stands for Pop. Your program should, given pairs of words produce sequences of stack operations which convert the first word to the second.

A stack is a data storage and retrieval structure permitting two operations:

Push - to insert an item and
Pop - to retrieve the most recently pushed item
We will use the symbol i (in) for push and o (out) for pop operations for an initially empty stack of characters. Given an input word, some sequences of push and pop operations are valid in that every character of the word is both pushed and popped, and furthermore, no attempt is ever made to pop the empty stack. For example, if the word FOO is input, then the sequence:

i i o i o o is valid, but
i i o is not (it's too short), neither is
i i o o o i (there's an illegal pop of an empty stack)

Valid sequences yield rearrangements of the letters in an input word. For example, the input word FOO and the sequence i i o i o o produce the anagram OOF. So also would the sequence i i i o o o. You are to write a program to input pairs of words and output all the valid sequences of i and o which will produce the second member of each pair from the first.

 

Input

The input will consist of several lines of input. The first line of each pair of input lines is to be considered as a source word (which does not include the end-of-line character). The second line (again, not including the end-of-line character) of each pair is a target word. The end of input is marked by end of file.

 

Output

For each input pair, your program should produce a sorted list of valid sequences of i and o which produce the target word from the source word. Each list should be delimited by

[
]

and the sequences should be printed in "dictionary order". Within each sequence, each i and o is followed by a single space and each sequence is terminated by a new line.

 

Sample Input

     
     
     
     

      
      
      
      madam
adamm
bahama
bahama
long
short
eric
rice 
     
     
     
     

 

Sample Output

   
   
   
   

    
    
    
    [
i i i i o o o i o o 
i i i i o o o o i o 
i i o i o i o i o o 
i i o i o i o o i o 
]
[
i o i i i o o i i o o o 
i o i i i o o o i o i o 
i o i o i o i i i o o o 
i o i o i o i o i o i o 
]
[
]
[
i i o i o i o o 
]
        

这道题开始写出来了，只有两种状态，一个是入栈，一个是出栈，但是我在出栈的时候一直没回溯，导致
栈数据被更改，所以在
    
    
    
    出栈搜索完后需要返回原来的状态。
 

    
    
    
    
#include <stdio.h>
#include <string.h>
const int maxn = 1005;
char ch[maxn], str[maxn], stack[maxn], ans[maxn];
int len;
void print ( int n )
{
    for ( int i = 0; i < n; i ++ )
        printf ( "%c ", ans[i] );   //注意最后也有空格
    printf ( "\n" );
}
void dfs ( int k, int pos, int top, int cnt )
{
    if ( k >= len )
    {
        while ( top >= 0 && stack[top] == str[pos] )
        //最后需要将所有能出栈的全部出栈
        {
            ans[cnt ++] = 'o';
            top --;
            pos ++;
        }
        if ( pos >= len )   //str全部出栈后就打印
            print ( cnt );
        return ;
    }
    ans[cnt] = 'i';
    stack[top+1] = ch[k];
    //入栈,但是注意出栈后,入栈会将以前数据修改,所以出栈时必须回溯
    dfs ( k+1, pos, top+1, cnt+1 );
    if ( top >= 0 && stack[top] == str[pos] )
    {
        ans[cnt] = 'o';
        char c = stack[top];    //保存栈顶的值
        dfs ( k, pos+1, top-1, cnt+1 );
        //因为top-1,入栈会将数据修改 并且不会恢复,入栈只有一次
        stack[top] = c; //恢复原来栈顶的值
    }
}
int main ( )
{
    int l;
    while ( ~ scanf ( "%s%s", ch, str ) )
    {
        len = strlen ( ch );
        l = strlen ( str );
        printf ( "[\n" );
        if ( l == len ) //剪掉长度不同的序列
            dfs ( 0, 0, -1, 0 );
        printf ( "]\n" );
    }
    return 0;
}