2015 多校联赛 ——HDU5414（）

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

题意:

每组测试数据输入两个字符串a和b，你可以选择字符串a中的任意一个字母x，在这个字母的后面加任意一个不是x的字母.

思路：

如果a比b长，不考虑

1. a 的序列是否能在 b中找到

2. a中前x个字母相同，b中前y个字母相同，  x >=y

aaab - > aabab 或者  aaab->aaabb

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 200050

char stra[100005];
char strb[100005];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int len1= 0,len2 = 0;
        scanf("%s%s",stra,strb);
        len1 = strlen(stra);
        len2 = strlen(strb);

        if(len1 > len2)
            printf("No
");
        else
        {
            int flag = 1,i = 0,j = 0;
            for(; j<len2&&i<len1; j++)    //a是否是b的子串
                if(strb[j]==stra[i]) i++;
            if(i<len1) flag = 0;
            i=0,j=0;
            while(i<len1&&stra[i]==stra[0])  //aaab前面连续相同字母不可能变多
                i++;
            while(j<len2&&strb[j]==strb[0])
                j++;
            if(i >= j && flag  && stra[0] == strb[0])
                printf("Yes
");
            else
                printf("No
");
        }
    }
    return 0;
}