浙大ZOJ 1004 Anagrams by Stack问题解决

继续分享代码。

一、工程代码及算法设计注释

--------------------------------------------------anagrams_by_stack.h----------------------------------------------

#ifndef JLU_CCST_GDC_CRASH_ANAGRAMS_BY_STACK#define JLU_CCST_GDC_CRASH_ANAGRAMS_BY_STACK#include <string>extern int findStackSequences(int str1Pos,int str2Pos,std::string str1,std::string str2);extern void testFindStackSequences();#endif//JLU_CCST_GDC_CRASH_ANAGRAMS_BY_STACK

--------------------------------------------------anagrams_by_stack.cpp----------------------------------------------

/**题目来源：浙大ACM在线测试——ZOJ，题目编号1004，题名"Anagrams by Stack"URL：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=1004Author：hellogdc<[e-mail]>Finish Time:2013.12.31*/#include <iostream>#include <list>#include <stack>#include <string>using namespace std;/**算法设计：递归思想求解：对于一个字符串而言，从其第k个字符出发求解到达目标字符串的出栈、入栈队列，可以转化为2个子部分：1. 将第k个字符入栈：保存字符i，然后求解从第k+1个字符出发求解到达目标字符串的出栈、入栈队列。2. 出栈：判断出栈的字符是否与对应的目标字符串相应位置的字符相同。如相同，则说明该次出栈成功，则保存字符o，然后继续求解从第k个字符出发求解到达目标字符串的出栈、入栈队列；否则说明此路不通，返回。虽然该部分与”父部分“差不多，但是不会陷入死循环，因为栈元素有限，不可能无限出栈；还有目标字符串长度也有限，也会导致该步终会终止。3. 递归出口：k比初始字符串长度大或者已发生字符串匹配。注：先处理入栈，然后再处理出栈，这样保证了每个输出的栈序列都是按字典序递增的。因为入栈的是小字符'i'，出栈的是大字符'o'。*//**str1Pos:第1个字符串的当前字符位置str2Pos:第2个字符串的当前字符位置str1:第1个字符串str2:第2个字符串返回：满足条件的序列的个数*/static list<char> sequenceList;static stack<char> strStack;int findStackSequences(int str1Pos,int str2Pos,string str1,string str2){//检查参数合法性if(str1Pos<0||str1.length()<0||str1.length()<=0||str2.length()<=0)return 0;int count=0;if(str1Pos>str1.length()||str2Pos>str2.length())return 0;//说明找到了一个满足条件的栈序列if(str2Pos==str2.length()){for(list<char>::iterator it=sequenceList.begin();it!=sequenceList.end();it++){cout<<(*it)<<'\t';}cout<<endl;return 1;}//入栈//检查是否可以入栈if(str1Pos<str1.length()){strStack.push(str1[str1Pos]);sequenceList.push_back('i');count+=findStackSequences(++str1Pos,str2Pos,str1,str2);//还原现场strStack.pop();sequenceList.pop_back();str1Pos--;}//出栈//检查是否可以出栈if(!strStack.empty()){char top=strStack.top();if(top==str2[str2Pos]){sequenceList.push_back('o');strStack.pop();count+=findStackSequences(str1Pos,++str2Pos,str1,str2);//还原现场strStack.push(top);sequenceList.pop_back();str2Pos--;}}return count;}void testFindStackSequences(){int count=0;string str1="TROT";string str2="TORT";cout<<"*****************************************************************"<<endl;cout<<"string1="<<str1<<endl;cout<<"string2="<<str2<<endl;cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;cout<<"count="<<count<<endl;cout<<"*****************************************************************"<<endl;str1="madam";str2="adamm";cout<<"string1="<<str1<<endl;cout<<"string2="<<str2<<endl;cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;cout<<"count="<<count<<endl;cout<<"*****************************************************************"<<endl;str1="bahama";str2="bahama";cout<<"string1="<<str1<<endl;cout<<"string2="<<str2<<endl;cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;cout<<"count="<<count<<endl;cout<<"*****************************************************************"<<endl;str1="long";str2="short";cout<<"string1="<<str1<<endl;cout<<"string2="<<str2<<endl;cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;cout<<"count="<<count<<endl;cout<<"*****************************************************************"<<endl;str1="eric";str2="rice";cout<<"string1="<<str1<<endl;cout<<"string2="<<str2<<endl;cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;cout<<"count="<<count<<endl;cout<<"*****************************************************************"<<endl;}

--------------------------------------------------main.cpp----------------------------------------------

#if 1#include "anagrams_by_stack.h"int main(){testFindStackSequences();return 0;}#endif

二、提交并被ZOJ成功接受的代码——算法核心代码

--------------------------------------------------submit_main.cpp----------------------------------------------

/**all the code are copyed from anagrams_by_stack.cpp*/#include <iostream>#include <list>#include <stack>#include <string>using namespace std;static list<char> sequenceList;static stack<char> strStack;static int findStackSequences(int str1Pos,int str2Pos,string str1,string str2){if(str1Pos<0||str1.length()<0||str1.length()<=0||str2.length()<=0)return 0;int count=0;if(str1Pos>str1.length()||str2Pos>str2.length())return 0;if(str2Pos==str2.length()){for(list<char>::iterator it=sequenceList.begin();it!=sequenceList.end();it++){cout<<(*it)<<' ';}cout<<endl;return 1;}if(str1Pos<str1.length()){strStack.push(str1[str1Pos]);sequenceList.push_back('i');count+=findStackSequences(++str1Pos,str2Pos,str1,str2);strStack.pop();sequenceList.pop_back();str1Pos--;}if(!strStack.empty()){char top=strStack.top();if(top==str2[str2Pos]){sequenceList.push_back('o');strStack.pop();count+=findStackSequences(str1Pos,++str2Pos,str1,str2);strStack.push(top);sequenceList.pop_back();str2Pos--;}}return count;}int main(){int count=0;string str1="TROT";string str2="TORT";while(cin>>str1>>str2){cout<<'['<<endl;count=findStackSequences(0,0,str1,str2);cout<<']'<<endl;}return 0;}