CCPC2018-湖南全国邀请赛 G String Transformation

G.String Transformation

题目描述

Bobo has a string S = s1 s2...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 0 = u ? v and vice versa where u , v are (possibly empty) strings and w ∈ { aa , bb , abab } .
Given the target string T = t1 t2 . . . tm , determine if Bobo can transform the string S into T .

输入

The input consists of several test cases and is terminated by end-of-file.
The first line of each test case contains a string s1 s2 ...sn . The second line contains a string t1 t2 . . . tm .

输出

For each test case, print Yes if Bobo can. Print No otherwise.
• 1 ≤ n, m ≤ 10⁴
• s1 , s2 ,..., sn , t1 , t2 , . . . , tm ∈ { a , b , c }
• The sum of n and m does not exceed 250,000.

样例输入

ab
ba
ac
ca
a
ab

样例输出

Yes
No
No

提示

For the first sample, Bobo can transform as ab => aababb => babb => ba .

题意：给定字符串S和T(均只由a,b,c三个字母组成)，可对S进行插入或删除操作(插入或删除的子串只能是"aa","bb"或"abab")，问能否通过操作将S变为T

 

     麻烦的模拟题，不管多麻烦的ababa的式子（不含c时）最终只会化成三种形式：a，b，ab（或ba），题目中插入删除abab的意义所在就是告诉我们ab能够转化成ba

    我的思路：将两个式子化解成最优。

eg：abaabcaacabc  化简 accabc，化简使用队列，边放进去边判，队末尾是a，要放的也是a，那就pop（a）

        ababaccbac      化简 accbac

然后在从队首把他们取出来进行比较，特判一下ab和ba是一样的。

#include <iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<deque>
#define ll long long
using namespace std;
char a[];
char b[];
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=;
        int c2=;
        for(int i=;i<la;i++)
        {
            if(a[i]=='c')
            {
                q1.push_back();
                c1++;
            }
            else if(a[i]=='a')
            {
                if(q1.empty ())
                {
                    q1.push_back();
                    continue;
                }
                if(q1.back()==)
                {
                    q1.pop_back();
                }
                else if(q1.back()==)
                {
                    if(i<la-&&a[i+]=='b')
                    {
                        q1.pop_back();
                        i++;
                        if(q1.empty ()||q1.back ()!=)
                        {
                            q1.push_back ();
                        }
                        else if(q1.back ()==)
                        {
                            q1.pop_back();
                        }

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

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

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

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

            }
            else if(b[i]=='b')
            {
                if(q2.empty ())
                {
                    q2.push_back();
                    continue;
                }
                if(q2.back()==)
                {
                    q2.pop_back();
                }
                else if(q2.back()==)
                {
                    if(i<lb-&&b[i+]=='a')
                    {
                        q2.pop_back();
                        i++;
                        if(q2.empty ()||q2.back ()!=)
                        {
                            q2.push_back ();
                        }
                        else if(q2.back ()==)
                        {
                            q2.pop_back();
                        }
                    }
                    else q2.push_back();

                }
                else q2.push_back();
            }
        }
        bool f=;
        if(c1!=c2||q1.size()!=q2.size()) f=;
        else
        {
            while(!q1.empty ())
            {
                if(q1.front() ==&&q2.front() ==)
                {
                    q1.pop_front ();
                    q2.pop_front ();
                    if(q1.empty ())
                    {
                        f=;
                        break;
                    }
                    else if(q1.front() ==&&q2.front() ==)
                    {
                        q1.pop_front ();
                        q2.pop_front ();
                    }
                    else
                    {
                        f=;
                        break;
                    }
                }
                else if(q1.front() ==&&q2.front() ==)
                {
                    q1.pop_front ();
                    q2.pop_front ();
                    if(q1.empty ())
                    {
                        f=;
                        break;
                    }
                    else if(q1.front() ==&&q2.front() ==)
                    {
                        q1.pop_front ();
                        q2.pop_front ();
                    }
                    else
                    {
                        f=;
                        break;
                    }
                }
                else if(q1.front()==q2.front ())
                {
                    q1.pop_front();
                    q2.pop_front();
                }
                else
                {
                    f=;
                    break;
                }
            }
        }
        if(f) printf("Yes\n");
        else printf("No\n");
    }
    return ;
}