HDOJ 5414 CRB and String 模拟

Posted 2020-09-22 blfbuaa

CRB and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 491    Accepted Submission(s): 186

Problem Description

CRB has two strings s and t.
In each step, CRB can select arbitrary character c of s and insert any character d (d ≠ c) just after it.
CRB wants to convert s to t. But is it possible?

 

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case there are two strings s and t, one per line.
1 ≤ T ≤ 105
1 ≤ |s| ≤ |t| ≤ 105
All strings consist only of lowercase English letters.
The size of each input file will be less than 5MB.

 

Output

For each test case, output "Yes" if CRB can convert s to t, otherwise output "No".

 

Sample Input

4
a
b
cat
cats
do
do
apple
aapple

 

Sample Output

No
Yes
Yes
No

 

Author

KUT（DPRK）

 

Source

2015 Multi-University Training Contest 10

 

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月21日 星期五 09时23分23秒
File Name     :HDOJ5414.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

const int maxn=100100;

int n,m;
int pip[maxn];
char S[maxn],T[maxn];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%s",S);
        scanf("%s",T);
        n=strlen(S); m=strlen(T);
        if(n>m) { puts("No"); }
        else if(n==m)
        {
            if(strcmp(S,T)==0) puts("Yes");
            else puts("No");
        }
        else
        {
            T[m]='&';
            bool flag=true;
			if(S[0]!=T[0]) flag=false;
            for(int i=0;i<n&&flag;i++)
            {
                bool temp=false;
                int st=0;
                if(i) st=pip[i-1]+1;
                for(int j=st;j<m;j++)
                {
                    if(S[i]==T[j]&&S[i]!=T[j+1])
                    {
                        /// find a point
                        pip[i]=j;
                        /// go back
                        int ni=i+1;
                        int ed=0;
                        if(i) ed=pip[i-1]+1;
                        for(int k=pip[i]-1;k>=ed;k--)
                        {
                            if(T[k]==S[ni]) 
			    {
				pip[ni]=pip[i];
				ni++;
			    }
                            else break;
                        }
                        i=ni-1;
                        temp=true; break;
                    }
                }
                if(temp==false) flag=false;
            }

            /// spc judge start point
            if(pip[0]!=0&&flag)
            {
                int len=1;
                for(int i=1;i<n;i++)
                {
                    if(S[i]==S[0]&&pip[0]==pip[i]&&len<pip[0]+1) len++;
                    else break;
                }
            	if(len!=pip[0]+1) flag=false;
            }

            if(flag==true) puts("Yes");
            else puts("No");
        }
    }
    return 0;
}