HDU

Anagrams by Stack

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2220    Accepted Submission(s): 1030

Problem 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

madamadammbahamabahamalongshortericrice

 

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 ]

题意：

        输入两个字符串，通过入栈出栈的操作，使出栈元素顺序满足第二个字符串。按字典序输出所有可能的操作顺序（i表示入栈，o表示出栈）。

题解：

        dfs+stack。搜索第一个字符串，一个字符有两种次状态，入栈或出栈。注意返回时要进行相反操作。

#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>#include<iostream>using namespace std;char str1[10010],str2[10010],buf[20010];int len;stack<char> s;void dfs(int in,int out,int step){    if(out==len)    {        for(int i=0;i<step;i++)        {            printf("%c ",buf[i]);        }        printf("\n");        return;    }    if(in<len)    {        buf[step]='i';        s.push(str1[in]);        dfs(in+1,out,step+1);        s.pop();    }    if(!s.empty() && str2[out]==s.top())    {        buf[step]='o';        s.pop();        dfs(in,out+1,step+1);        s.push(str2[out]);    }}int main(){    while(~scanf("%s",str1))    {        scanf("%s",str2);        while(!s.empty())            s.pop();        len=strlen(str1);        printf("[\n");        dfs(0,0,0);        printf("]\n");    }    return 0;}