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…

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…

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 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…

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1004 题目大意:给你个栈,给你源串和目标串,按字典序输出符合要求的进站出站序列. 就是搜搜搜呗... 带上答案和模拟的栈.. 代码: #include <cstdio> #include <cstdlib> #include <string> #include <iostream> #include <cstring&…

考察DFS的应用,用栈描述字符串的变化过程. #include <stdio.h> #include <string.h> int len1,len2; ],str2[],stk[],ans[]; void output(int n){ int i; ;i<n;i++){ printf("%c ",ans[i]); } printf("\n"); } void go(int in,int out,int p,int top){ ) re…

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…

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…

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…

细节问题各种虐!! 其实就是简单的一个深搜 看成二叉树来理解:每个节点有两个枝:入栈和出栈. 剪枝操作:只有当栈顶元素和当前位置的目标字符相同时才出栈,否则就不出栈 dfs写三个参数:depth搜索深度,npush压栈数,npop出栈数 npush用于记录压栈数:主要判断当前压栈是否合理,以及要压入的元素在原串种的位置 npop用于记录出栈数:判断生成的目标串元素的位置 当npush==npop==目标串时,说明生成了一个可执行的操作串 注意输出操作串的时候一定要用depth参数来控制,因为在多…