CodeForces - 1194C From S To T （模拟、水题）

向右看齐 2021-11-09 07:28 288阅读 0赞

You are given three strings ss, tt and pp consisting of lowercase Latin letters. You may perform any number (possibly, zero) operations on these strings.

During each operation you choose any character from pp, erase it from pp and insert it into string ss (you may insert this character anywhere you want: in the beginning of ss, in the end or between any two consecutive characters).

For example, if pp is aba, and ss is de, then the following outcomes are possible (the character we erase from pp and insert into ss is highlighted):

*  aba →→ ba, de →→ ade;
 *  aba →→ ba, de →→ dae;
 *  aba →→ ba, de →→ dea;
 *  aba →→ aa, de →→ bde;
 *  aba →→ aa, de →→ dbe;
 *  aba →→ aa, de →→ deb;
 *  aba →→ ab, de →→ ade;
 *  aba →→ ab, de →→ dae;
 *  aba →→ ab, de →→ dea;

Your goal is to perform several (maybe zero) operations so that ss becomes equal to tt. Please determine whether it is possible.

Note that you have to answer qq independent queries.

Input

The first line contains one integer qq (1≤q≤1001≤q≤100) — the number of queries. Each query is represented by three consecutive lines.

The first line of each query contains the string ss (1≤|s|≤1001≤|s|≤100) consisting of lowercase Latin letters.

The second line of each query contains the string tt (1≤|t|≤1001≤|t|≤100) consisting of lowercase Latin letters.

The third line of each query contains the string pp (1≤|p|≤1001≤|p|≤100) consisting of lowercase Latin letters.

Output

For each query print YES if it is possible to make ss equal to tt, and NO otherwise.

You may print every letter in any case you want (so, for example, the strings yEs, yes, Yes and YES will all be recognized as positive answer).

Input

4
    ab
    acxb
    cax
    a
    aaaa
    aaabbcc
    a
    aaaa
    aabbcc
    ab
    baaa
    aaaaa

Output

YES
    YES
    NO
    NO

Note

In the first test case there is the following sequence of operation:

1.  s=s= ab, t=t= acxb, p=p= cax;
2.  s=s= acb, t=t= acxb, p=p= ax;
3.  s=s= acxb, t=t= acxb, p=p= a.

In the second test case there is the following sequence of operation:

1.  s=s= a, t=t= aaaa, p=p= aaabbcc;
2.  s=s= aa, t=t= aaaa, p=p= aabbcc;
3.  s=s= aaa, t=t= aaaa, p=p= abbcc;
4.  s=s= aaaa, t=t= aaaa, p=p= bbcc.

这里贴一个别人的代码

#include<bits/stdc++.h>
    using namespace std;
    #define forn(i,n)  for(int i=0;i<n;i++)
    #define rep(i,a,n) for(int i=a;i<=n;i++)
    const int maxn=1e5+5;  
    const int INF=0x3f3f3f3f;
    
    
    int a[maxn],sum[maxn],c[maxn];
    int n,q;
    string s,t,p;
    int solve(){
        int i=0,j=0;
        int lens=s.size(),lent=t.size();
        while(i<lens && j<lent){
            if(s[i]==t[j])
                i++;
            else
                c[t[j]-'a']++;  
            j++;
        }
        if(i<lens)    
        {
            return 0;
        } 
        if(j<lent){
            for(;j<lent;j++){        
                c[t[j]-'a']++;
            }
        }    
        for(char o:p){ 
            c[o-'a']--;
        }
        for(int i=0;i<26;i++){
            if(c[i]>0)
                return 0;
        }
        return 1;
    } 
    int main(){
        
        
        cin>>q;
        while(q--){
            cin>>s>>t>>p;   
            memset(c,0,sizeof c);
            cout<<(solve()?"YES":"NO")<<endl;
        }
        
        return 0;
    }

2019-7-24

发现了昨天双指针发的错误， 没有判断第一个串有没有读尽，

哎，是我愚蠢了

#include<bits/stdc++.h>
    using namespace std;
    int ss[26], tt[26], need[26], have[26];
    #define ZERO(a) memset(a, 0, sizeof(a))
    
    int main()
    {
        int T;
        cin >> T;
        while(T--)
        {
            string s,t,p;
            cin >> s >> t >> p;
    
            ZERO(need); ZERO(have);
    
            for(int i = 0; i < p.size(); i++)
                have[p[i] - 'a']++;
    
            int j = 0;
            int flag = 0;
            for(int i = 0; i < t.size(); i++)
            {
                if(s[j] == t[i] && j < s.size()) j++;
                else
                {
                    if(have[t[i] - 'a'] > 0) --have[t[i] - 'a'];
                    else
                    {
                        flag = 1;
                        break;
                    }
                }
            }
            if(j != s.size()) flag = 1;
            if(flag == 0) puts("YES");
            else puts("NO");
        }
    
    
    }

转载于:https://www.cnblogs.com/YY666/p/11233438.html