hdu 1515 Anagrams by Stack

Anagrams by Stack

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 843 Accepted Submission(s): 407

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 ]

Source

Zhejiang University Local Contest 2001

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题目大意。模拟进出栈用给定序列得到目标序列。输出满足条件的方法。

#include <iostream>#include<stdio.h>#include<string.h>#include<stack>using namespace std;stack<char> sta;//用标准库建立一个栈char s1[100],s2[100],way[200];//s1存给定序列s2存目标序列。way记录操作int cnt;void print()//如果搜索到满足条件的方法则输出{    int i;    for(i=0; i<cnt; i++)        printf("%c ",way[i]);//末尾空格处理了还错了。晕！    printf("\n");}void dfs(int p1,int p2)//p1表示s1待入栈位置.p2表示s2待匹配位置{    if(s2[p2]=='\0')//如果目标序列已完全匹配输出答案返回    {        print();        return;    }    char c;    if(s1[p1]!='\0')//如果s1还没有完全进栈按题目要求字典顺序（先进栈后出栈搜索）    {        c=s1[p1];        sta.push(c);        way[cnt++]='i';        dfs(p1+1,p2);        sta.pop();        cnt--;    }    if(!sta.empty())//后出栈（如果栈不为空）开始没判断栈空，结果找了很久bug    {        c=sta.top();        if(c==s2[p2])//注意该处剪枝        {            sta.pop();            way[cnt++]='o';            dfs(p1,p2+1);            cnt--;            sta.push(c);        }    }}int main(){    while(~scanf("%s%s",s1,s2))    {        while(!sta.empty())            sta.pop();        cnt=0;        printf("[\n");        dfs(0,0);        printf("]\n");    }    return 0;}