CRB and String

最新推荐文章于 2018-08-16 11:43:45 发布

weixin_30411997

最新推荐文章于 2018-08-16 11:43:45 发布

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原文链接：http://www.cnblogs.com/juechen/p/5255958.html

本文深入探讨了一种独特的字符串操作方法——通过在选定字符后插入非匹配字符，实现从初始字符串到目标字符串的转换。通过具体实例和逻辑分析，展示了如何判断并实施这一过程，以验证是否可能将给定的初始字符串转化为指定的目标字符串。

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CRB and String
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 543 Accepted Submission(s): 208

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）
题意:给你两个字符串s和t，你可以在字符串s中任意选一个字符c，在该字符c后插入一个字符d(d!=c)，问经过多次此操作，能否将字符串s转化成字符串t;
思路:
对于这两个字符串,要能够插入成功必须符合两个条件
1:s是t的字串
2:对于t前k个字符如果是相同的s的前k个字符也必须是相同的,否则无法插入(想想为什么??)
只要符合上面的条件就都可以转化成功

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std;

typedef unsigned long long LL;

const int Max = 110000;

char s[Max],t[Max];

bool flag;

int main()
{
    int T;
    int i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        scanf("%s",t);
        flag=false;
        i=0,j=0;
        for(; t[i]!='\0'; i++)
        {
            if(!flag&&t[i]==t[0]&&s[j]!=t[i])
            {
                break;
            }
            if(t[i]!=t[0])
            {
                flag=true;
            }
            if(s[j]!='\0'&&t[i]==s[j])
            {
                j++;
            }
        }
        if(t[i]=='\0'&&s[j]=='\0')
        {
            printf("Yes\n");
        }
        else
        {
            printf("No\n");
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/juechen/p/5255958.html