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A Great Alchemist

系統 1801 0

Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MB

Problem
Carol is a great alchemist.

In her world, each metal has a name of 2N (N is an integer) letters long, which consists of uppercase alphabets.

Carol can create metal S3 from S1 and S2 alchemical when she can make the name of S3 by taking N letters each from S1 and S2 then rearranging them properly.

You are given 3 names of the metal S1, S2, S3. Determine wether Carol can create S3 from S1 and S2 or not.

Input
The input will be given in the following format from the Standard Input.

S1
S2
S3
On the first line, you will be given the name of the first metal material S1.
On the second line, you will be given the name of the second metal material S2.
On the third line, you will be given the name of the metal S3, which Carol wants to create.
Each character in the S1, S2, and S3 will be an uppercase English alphabet letter.
Each string S1, S2 and S3 has same number of letters and the number is always even.
It is guaranteed that 2≦|S1|≦100000
Output
If Carol can create S3 from S1 and S2, output YES, if not, output NO in one line. Make sure to insert a line break at the end of the output.

Input Example 1
AABCCD
ABEDDA
EDDAAA
Output Example 1
YES
You can make EDDAAA by picking AAD from the first metal, and AED from the second metal.

Input Example 2
AAAAAB
CCCCCB
AAABCB
Output Example 2
NO
To make AAABCB, you have to take at least four letters from the first material. So this can't be created alchemical.

用回溯法TLE。看了同學的代碼,在執行回溯前執行一些檢查就能過了。哎。

      
         1
      
       #include <iostream>


      
         2
      
       #include <
      
        string
      
      >


      
         3
      
       #include <vector>


      
         4
      
      
        using
      
      
        namespace
      
      
         std;


      
      
         5
      
      
         6
      
      
        bool
      
       backtrack(
      
        string
      
       &S3, 
      
        int
      
       charsFromS1, 
      
        int
      
       charsFromS2, 
      
        int
      
      
         current, 


      
      
         7
      
               vector<
      
        int
      
      > &charsInS1, vector<
      
        int
      
      > &
      
        charsInS2) {


      
      
         8
      
      
        if
      
       (current >= S3.length()) 
      
        return
      
      
        true
      
      
        ;


      
      
         9
      
      
        char
      
       index = S3[current] - 
      
        '
      
      
        A
      
      
        '
      
      
        ;


      
      
        10
      
      
        if
      
       (charsInS1[index] > 
      
        0
      
       && charsFromS1 < S3.length() / 
      
        2
      
      
        ) {


      
      
        11
      
               charsInS1[index]--
      
        ;


      
      
        12
      
      
        if
      
       (backtrack(S3, charsFromS1 + 
      
        1
      
      , charsFromS2, current + 
      
        1
      
      , charsInS1, charsInS2)) 
      
        return
      
      
        true
      
      
        ;


      
      
        13
      
               charsInS1[index]++
      
        ;


      
      
        14
      
      
            }


      
      
        15
      
      
        if
      
       (charsInS2[index] > 
      
        0
      
       && charsFromS2 < S3.length() / 
      
        2
      
      
        ) {


      
      
        16
      
               charsInS2[index]--
      
        ;


      
      
        17
      
      
        if
      
       (backtrack(S3, charsFromS1, charsFromS2 + 
      
        1
      
      , current + 
      
        1
      
      , charsInS1, charsInS2)) 
      
        return
      
      
        true
      
      
        ;


      
      
        18
      
               charsInS2[index]++
      
        ;


      
      
        19
      
      
            }


      
      
        20
      
      
        return
      
      
        false
      
      
        ;


      
      
        21
      
      
        }


      
      
        22
      
      
        23
      
      
        int
      
       main(
      
        int
      
       argc, 
      
        char
      
      **
      
         argv) {


      
      
        24
      
      
        string
      
      
         S1, S2, S3;


      
      
        25
      
           cin >> S1 >> S2 >>
      
         S3;


      
      
        26
      
           vector<
      
        int
      
      > charsInS1(
      
        26
      
      , 
      
        0
      
      ), charsInS2(
      
        26
      
      , 
      
        0
      
      ), charsInS3(
      
        26
      
      , 
      
        0
      
      
        );


      
      
        27
      
      
        28
      
      
        for
      
       (
      
        int
      
       i = 
      
        0
      
      ; i < S1.length(); ++
      
        i) {


      
      
        29
      
               charsInS1[S1[i] - 
      
        '
      
      
        A
      
      
        '
      
      ]++
      
        ;


      
      
        30
      
               charsInS2[S2[i] - 
      
        '
      
      
        A
      
      
        '
      
      ]++
      
        ;


      
      
        31
      
               charsInS3[S3[i] - 
      
        '
      
      
        A
      
      
        '
      
      ]++
      
        ;


      
      
        32
      
      
            }


      
      
        33
      
      
        34
      
      
        int
      
       common13 = 
      
        0
      
      , common23 = 
      
        0
      
      
        ;


      
      
        35
      
      
        for
      
       (
      
        int
      
       i = 
      
        0
      
      ; i < 
      
        26
      
      ; ++
      
        i) {


      
      
        36
      
      
        if
      
       (charsInS3[i] > charsInS1[i] +
      
         charsInS2[i]) {


      
      
        37
      
                   cout << 
      
        "
      
      
        NO
      
      
        "
      
       <<
      
         endl;


      
      
        38
      
      
        return
      
      
        0
      
      
        ;


      
      
        39
      
      
                }


      
      
        40
      
               common13 +=
      
         min(charsInS3[i], charsInS1[i]);


      
      
        41
      
               common23 +=
      
         min(charsInS3[i], charsInS2[i]);


      
      
        42
      
      
            }


      
      
        43
      
      
        44
      
      
        if
      
       (common13 < S3.length() / 
      
        2
      
       || common23 < S3.length() / 
      
        2
      
      
        ) {


      
      
        45
      
               cout << 
      
        "
      
      
        NO
      
      
        "
      
       <<
      
         endl;


      
      
        46
      
           } 
      
        else
      
      
         {


      
      
        47
      
      
        bool
      
       ans = backtrack(S3, 
      
        0
      
      , 
      
        0
      
      , 
      
        0
      
      
        , charsInS1, charsInS2);


      
      
        48
      
               cout << (ans ? 
      
        "
      
      
        YES
      
      
        "
      
       : 
      
        "
      
      
        NO
      
      
        "
      
      ) <<
      
         endl;


      
      
        49
      
      
            }


      
      
        50
      
      
        return
      
      
        0
      
      
        ;


      
      
        51
      
       }
    

?

A Great Alchemist


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