iostream ear out delet eve 麻煩 感謝 back 所在

String Transformation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 36 Accepted Submission(s): 8

Problem Description Bobo has a string S=s1s2…sn consists of letter `a`, `b` and `c`.
He can transform the string by inserting or deleting substrings `aa`, `bb` and `abab`.

Formally, A=u°w°v

(``° ‘‘ denotes string concatenation) can be transformed into A′=u°v and vice versa where u , v are (possibly empty) strings and w∈{aa,bb,abab} .

Given the target string T=t1t2…tm , determine if Bobo can transform the string S into T .

Input The input consists of several test cases and is terminated by end-of-file.

The first line of each test case contains a string s1s2…sn

.
The second line contains a string t1t2…tm .

Output For each test case, print `Yes` if Bobo can. Print `No` otherwise.

## Constraint

* 1≤n,m≤104
* s1,s2,…,sn,t1,t2,…,tm∈{a,b,c}
* The sum of n and m does not exceed 250,000 .

Sample Input ab ba ac ca a ab

Sample Output Yes No No Hint For the first sample, Bobo can transform as `ab => aababb => babb => ba`.

Source CCPC2018-湖南全國邀請賽-重現賽（感謝湘潭大學） 麻煩的模擬題，不管多麻煩的ababa的式子（不含c時）最終只會化成三種形式：a，b，ab（或ba），題目中插入刪除abab的意義所在就是告訴我們ab能夠轉化成ba 我的思路：將兩個式子化解成最優。 eg：abaabcaacabc 化簡 accabc，化簡使用隊列，邊放進去邊判，隊末尾是a，要放的也是a，那就pop（a） ababaccbac 化簡 accbac

然後在從隊首把他們取出來進行比較，特判一下ab和ba是一樣的。

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#define ll long long
using namespace std;
char a[10005];
char b[10005];
deque<int>q1,q2;
int main()
{
    while(cin>>a>>b)
    {
        while(!q1.empty ()) q1.pop_back();
        while(!q2.empty ()) q2.pop_back();
        int la=strlen(a);
        int lb=strlen(b);
        int c1=0;
        int c2=0;
        for(int i=0;i<la;i++)
        {
            if(a[i]==‘c‘)
            {
                q1.push_back(3);
                c1++;
            }
            else if(a[i]==‘a‘)
            {
                if(q1.empty ()) 
                {
                    q1.push_back(1);
                    continue;
                }
                if(q1.back()==1)
                {
                    q1.pop_back();
                }
                else if(q1.back()==2)
                {
                    if(i<la-1&&a[i+1]==‘b‘)
                    {
                        q1.pop_back();
                        i++;
                        if(q1.empty ()||q1.back ()!=1) 
                        {
                            q1.push_back (1);
                        }
                        else if(q1.back ()==1)
                        {
                            q1.pop_back();
                        }

                    }
                    else q1.push_back (1);
                }
                else q1.push_back (1);

            }
            else if(a[i]==‘b‘)
            {
                if(q1.empty ()) 
                {
                    q1.push_back(2);
                    continue;
                }
                if(q1.back()==2)
                {
                    q1.pop_back();
                }
                else if(q1.back()==1)
                {
                    if(i<la-1&&a[i+1]==‘a‘)
                    {
                        q1.pop_back();
                        i++;
                        if(q1.empty ()||q1.back ()!=2) 
                        {
                            q1.push_back (2);
                        }
                        else if(q1.back ()==2)
                        {
                            q1.pop_back();
                        }

                    }
                    else q1.push_back(2);
                }
                else q1.push_back(2);
            }
        }
        for(int i=0;i<lb;i++)
        {
            if(b[i]==‘c‘)
            {
                q2.push_back(3);
                c2++;
            }
            else if(b[i]==‘a‘)
            {
                if(q2.empty ()) 
                {
                    q2.push_back(1);
                    continue;
                }
                if(q2.back()==1)
                {
                    q2.pop_back();
                }
                else if(q2.back()==2)
                {
                    if(i<lb-1&&b[i+1]==‘b‘)
                    {
                        q2.pop_back();
                        i++;
                        if(q2.empty ()||q2.back ()!=1) 
                        {
                            q2.push_back (1);
                        }
                        else if(q2.back ()==1)
                        {
                            q2.pop_back();
                        }

                    }
                    else q2.push_back (1);
                }
                else q2.push_back (1);

            }
            else if(b[i]==‘b‘)
            {
                if(q2.empty ()) 
                {
                    q2.push_back(2);
                    continue;
                }
                if(q2.back()==2)
                {
                    q2.pop_back();
                }
                else if(q2.back()==1)
                {
                    if(i<lb-1&&b[i+1]==‘a‘)
                    {
                        q2.pop_back();
                        i++;
                        if(q2.empty ()||q2.back ()!=2) 
                        {
                            q2.push_back (2);
                        }
                        else if(q2.back ()==2)
                        {
                            q2.pop_back();
                        }
                    }
                    else q2.push_back(2);
                    
                }
                else q2.push_back(2);
            }
        }
        bool f=1;
        if(c1!=c2||q1.size()!=q2.size()) f=0;
        else
        {
            while(!q1.empty ())
            {
                if(q1.front() ==1&&q2.front() ==2)
                {
                    q1.pop_front ();
                    q2.pop_front ();
                    if(q1.empty ())
                    {
                        f=0;
                        break;
                    }
                    else if(q1.front() ==2&&q2.front() ==1)
                    {
                        q1.pop_front ();
                        q2.pop_front ();
                    }
                    else 
                    {
                        f=0;
                        break;
                    }
                }
                else if(q1.front() ==2&&q2.front() ==1)
                {
                    q1.pop_front ();
                    q2.pop_front ();
                    if(q1.empty ())
                    {
                        f=0;
                        break;
                    }
                    else if(q1.front() ==1&&q2.front() ==2)
                    {
                        q1.pop_front ();
                        q2.pop_front ();
                    }
                    else 
                    {
                        f=0;
                        break;
                    }
                }
                else if(q1.front()==q2.front ())
                {
                    q1.pop_front();
                    q2.pop_front();
                }
                else 
                {
                    f=0;
                    break;
                }
            }
        }
        if(f) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

CCPC2018-湖南全國邀請賽 G String Transformation