2159 -- Ancient Cipher

Ancient Cipher

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 36074		Accepted: 11765

Description

Ancient Roman empire had a strong government system with various departments, including a secret service department. Important documents were sent between provinces and the capital in encrypted form to prevent eavesdropping. The most popular ciphers in those times were so called substitution cipher and permutation cipher. 
Substitution cipher changes all occurrences of each letter to some other letter. Substitutes for all letters must be different. For some letters substitute letter may coincide with the original letter. For example, applying substitution cipher that changes all letters from 'A' to 'Y' to the next ones in the alphabet, and changes 'Z' to 'A', to the message "VICTORIOUS" one gets the message "WJDUPSJPVT". 
Permutation cipher applies some permutation to the letters of the message. For example, applying the permutation <2, 1, 5, 4, 3, 7, 6, 10, 9, 8> to the message "VICTORIOUS" one gets the message "IVOTCIRSUO". 
It was quickly noticed that being applied separately, both substitution cipher and permutation cipher were rather weak. But when being combined, they were strong enough for those times. Thus, the most important messages were first encrypted using substitution cipher, and then the result was encrypted using permutation cipher. Encrypting the message "VICTORIOUS" with the combination of the ciphers described above one gets the message "JWPUDJSTVP". 
Archeologists have recently found the message engraved on a stone plate. At the first glance it seemed completely meaningless, so it was suggested that the message was encrypted with some substitution and permutation ciphers. They have conjectured the possible text of the original message that was encrypted, and now they want to check their conjecture. They need a computer program to do it, so you have to write one.

Input

Input contains two lines. The first line contains the message engraved on the plate. Before encrypting, all spaces and punctuation marks were removed, so the encrypted message contains only capital letters of the English alphabet. The second line contains the original message that is conjectured to be encrypted in the message on the first line. It also contains only capital letters of the English alphabet. 
The lengths of both lines of the input are equal and do not exceed 100.

Output

Output "YES" if the message on the first line of the input file could be the result of encrypting the message on the second line, or "NO" in the other case.

Sample Input

JWPUDJSTVP
VICTORIOUS

Sample Output

YES

Source

Northeastern Europe 2004

 

题目大意：

古罗马帝王的保密服务部门的保密方法是替换和重新排列。

替换方法是将出现的字符替换成其他的字符。如将'A'替换成'Z'，将'Z'替换成'A'。

排列方法是改变原来单词中字母的顺序。例如将顺序变为<2,1,5,4,3,7,6,10,9,8>。应用到字符串

"VICTORIOUS"上，则可以得到"IVOTCIRSUO"。

单用一种解密方法是不安全的，只有将两种方法结合起来才安全。那么问题来了：给你一个原文

字符串和加密字符串，问是否能通过这两种加密方法结合，从而由原文信息得到加密信息。如果

能则输出"YES"，否则输出"NO"。

 

思路：

其实这道题没那么复杂，只要用两个数组分别存下两个字符串中各个字母的个数，排序一下，比较

字母个数是不是都相等就可以了，如果不是全相等，则说明不能从原文信息得到加密信息。如果全

相等，则一定有方法从原文信息得到加密信息。

#include<stdio.h>
#include<math.h>
#include<iostream>
#include<map>
using namespace std;
int main()
{
    map<char,int> First,Second;
    char c = getchar();
    while(c!='
')
    {
        First[c]++;
        c = getchar();
    }
    c = getchar();
    while(c!='
')
    {
        Second[c]++;
        c = getchar();
    }
    map<char,int>::iterator FIter = First.begin();
    for(;FIter!=First.end();)
    {
        int temp = 0;
        map<char,int>::iterator SIter = Second.begin();
        for(;SIter!=Second.end();SIter++)
        {
            if(FIter->second == SIter->second)
            {
                char c1,c2;
                c1 = FIter->first;c2 = SIter->first;
                FIter++;SIter = Second.begin();
                First.erase(c1);
                Second.erase(c2);
                temp = 1;
                break;
            }
        }
        if(!temp)
            FIter++;
    }
    //如果两个map不为空
    if(First.size()||Second.size())
    {
        cout<<"NO"<<endl;
    }else{
        cout<<"YES"<<endl;
    }
    First.clear();
    Second.clear();

    return 0;
}