zoj1004-Anagrams by Stack 【栈 dfs】】的更多相关文章

Anagrams by Stack 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4 题目大意:输入两个字符串序列,判断能否通过对第一个字符串进行进栈出栈操作得到第二个字符串,若能则输出所有能达到的进出栈操作过程.我通过全排列每得到一组操作过程,则用函数按照这个操作过程,判断能否得到第二个字符串,若能则表明此操作过程可行,输出. 代码如下: # include<iostream> # include<sta…
题目大意:规定 i 为入栈,o 为出栈,现在给两个字符串st1,st2,现在要将st1转化为st2,转化方法是,st1中字符从头开始入栈,并合理出栈构造出st2.请输出所有可能的出入栈步骤. 深度优先搜索+回溯~ #include<bits/stdc++.h> using namespace std; string s1,s2; int len; stack<char> st; vector<char> path; void dfs (int ipush,int ipo…
题目传送门 /* stack 容器的应用: 要求字典序升序输出,所以先搜索入栈的 然后逐个判断是否满足答案,若不满足,回溯继续搜索,输出所有符合的结果 */ #include <cstdio> #include <iostream> #include <algorithm> #include <stack> #include <cmath> #include <cstring> #include <vector> usin…
http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4 Anagrams by Stack Time Limit: 2 Seconds      Memory Limit: 65536 KB How can anagrams result from sequences of stack operations? There are two sequences of stack operators which can convert…
题目概述:Anagrams by Stack 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 progr…
Anagrams by Stack 题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1004 题意:通过堆栈实现将一个字符串转变成目标字符串的操作,要求输出全部的可能操作组合. 思路:利用深度优先的搜索思路,对于每一个状态都有入栈和出栈两种可能的操作,由于要求按字典序输出,每次先考虑入栈再考虑出栈.即"能入就入,不能入考虑是否能退,随后返回上一步". 下面贴代码: 1 //Problem Name: A…
Anagrams by Stack Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 870    Accepted Submission(s): 419 Problem Description How can anagrams result from sequences of stack operations? There are two…
ZOJ Problem Set - 1004 Anagrams by Stack Time Limit: 2 Seconds      Memory Limit: 65536 KB 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…
ZOJ Problem Set - 1004 Anagrams by Stack Time Limit: 2 Seconds      Memory Limit: 65536 KB 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…
首先分清楚Stack,Heap的中文翻译:Stack—栈,Heap—堆.         在中文里,Stack可以翻译为“堆栈”,所以我直接查找了计算机术语里面堆和栈开头的词语:        堆存储: heapstorage    堆存储分配: heapstorage allocation  堆存储管理: heap storage management         栈编址: stack addressing   栈变换:stack transformation  栈存储器:stack me…