hdu5414 CRB and String_rb has two strings s and t. in each step, crb can -优快云博客

本文介绍了一个字符串转换问题，即通过在给定字符串的任意字符后添加不同字符的方式，判断是否能将一个字符串转换为另一个字符串。文章提供了一种解题思路及实现代码。

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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：给你两个字符串s1,s2,你可以在s1中选择任意一个字符，并且在后面添加任意与之不同的字符在后面，如果添加若干或者不添加能使s1变成s2,就输出Yes,否则输出No.
思路：只要满足两个条件就输出Yes：1.两个字符串首字母必须相等且第一个字符串中首字母连续的个数必须小于等于第二个字符。2.对于任意一个字符c,它在第一个字符串中的个数要小于等于第二个字符串的。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define maxn 100050
char s1[maxn],s2[maxn];
int num1[27],num2[27];
int main()
{
    int n,m,i,j,T,len1,len2,flag1,sum1,sum2,flag2,flag;
    char c1,c2;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s%s",s1,s2);
        if(strcmp(s1,s2)==0){
            printf("Yes\n");continue;
        }
        len1=strlen(s1);
        len2=strlen(s2);
        c1=s1[0];c2=s2[0];
        if(c1!=c2){
            printf("No\n");continue;
        }
        memset(num1,0,sizeof(num1));
        memset(num2,0,sizeof(num2));
        sum1=sum2=0;flag1=flag2=1;
        for(i=1;i<len1;i++){
            if(flag1 && s1[i]==s1[0])sum1++;
            else flag1=0;
            num1[s1[i]-'a'+1]++;
        }
        for(i=1;i<len2;i++){
            if(flag2 && s2[i]==s2[0])sum2++;
            else flag2=0;
            num2[s2[i]-'a'+1]++;
        }
        if(sum1<sum2){
            printf("No\n");continue;
        }
        flag=1;
        for(i=1;i<=26;i++){
            if(num1[i]>num2[i]){
                flag=0;break;
            }
        }
        if(flag)printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}