hdu 5414 CRB and String(字符串模拟)

CRB and String

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1787    Accepted Submission(s): 579

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

4abcatcatsdodoappleaapple

 

Sample Output

NoYesYesNo

题意：有两个串s和t，问能不能通过往s串里多次插入字符从而使s串变成t串，注意插入的字符不能和插入位置的前一个字符相同。比如abc和aabc就是不合法的

思路：直接模拟便是。 注意abc和abbbbbbbbbbc是合法的，因为可以在a后面多次插入b

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>using namespace std;#define N 100005char s[N],t[N];int main(){    int T;    scanf("%d",&T);    while(T--)    {        scanf("%s %s",s,t);        int lens=strlen(s),lent=strlen(t),nt=1;        int flag=0;        if(s[0]!=t[0]) flag=1;        else        {            for(int i=1; i<lens; )            {                if(s[i]==t[nt])                {                    i++;                    nt++;                    continue;                }                int now=nt-1;                while(t[now]==t[nt]) now--;                if(now==-1)                {                    flag=1;                    break;                }                nt++;                if(nt>=lent)                {                    flag=1;                    break;                }            }            for(int i=nt; i<lent; i++)            {                if(t[i]==t[i-1])                {                    int now=nt-1;                    while(t[now]==t[nt]) now--;                    if(now==-1)                    {                        flag=1;                        break;                    }                }            }        }        if(flag) printf("No\n");            else printf("Yes\n");    }    return 0;}