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[ACM] POJ 1936 All in All （查找，一个串是否在另一个串中）

2024-07-12 14:29:58   224人阅读

All in All
Time Limit: 1000MS Memory Limit: 30000K
Total Submissions: 27521 Accepted: 11266

Description

You have devised a new encryption technique which encodes a message by inserting between its characters randomly generated strings in a clever way. Because of pending patent issues we will not discuss in detail how the strings are generated and inserted into the original message. To validate your method, however, it is necessary to write a program that checks if the message is really encoded in the final string.

Given two strings s and t, you have to decide whether s is a subsequence of t, i.e. if you can remove characters from t such that the concatenation of the remaining characters is s.

Input

The input contains several testcases. Each is specified by two strings s, t of alphanumeric ASCII characters separated by whitespace.The length of s and t will no more than 100000.

Output

For each test case output "Yes", if s is a subsequence of t,otherwise output "No".

Sample Input

sequence subsequence
person compression
VERDI vivaVittorioEmanueleReDiItalia
caseDoesMatter CaseDoesMatter

Sample Output

Yes
No
Yes
No

Source

Ulm Local 2002
代码：

一开始想多了，想着要查找的串的第一个字母对应另一个串中的很多位置怎么办。。。其实不用只要匹配了一个字母待匹配的串就少了一个字母，不需要回溯。。所以直接设置两个指针直接模拟就可以了。。

代码：

#include <iostream>
#include <string.h>
using namespace std;
string s1,s2;
int len1,len2;

int main()
{
    while(cin>>s1>>s2)
    {
        int len1=s1.length();
        int len2=s2.length();
        int i=0,j=0;
        while(j<len2)
        {
            if(s1[i]==s2[j])
            {
                i++;
                j++;
                if(i==len1)
                {
                    cout<<"Yes"<<endl;
                    break;
                }
            }
            else
            {
                 j++;
                 if(j==len2)
                 {
                     cout<<"No"<<endl;
                     break;
                 }
            }
        }
    }
    return 0;
}



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