Anagram

Posted 2020-11-08 zangzang

问题 A: Anagram

时间限制: 1 Sec  内存限制: 128 MB
提交: 89  解决: 53
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题目描述

Orz has two strings of the same length: A and B. Now she wants to transform A into an anagram of B (which means, a rearrangement of B) by changing some of its letters. The only operation the girl can make is to “increase” some (possibly none or all) characters in A. E.g., she can change an ‘A‘ to a ‘B‘, or a ‘K‘ to an ‘L‘. She can increase any character any times. E.g., she can increment an ‘A‘ three times to get a ‘D‘. The increment is cyclic: if she increases a ‘Z‘, she gets an ‘A‘ again.

For example, she can transform "ELLY" to "KRIS" character by character by shifting ‘E‘ to ‘K‘ (6 operations), ‘L‘ to ‘R‘ (again 6 operations), the second ‘L‘ to ‘I‘ (23 operations, going from ‘Z‘ to ‘A‘ on the 15-th operation), and finally ‘Y‘ to ‘S‘ (20 operations, again cyclically going from ‘Z‘ to ‘A‘ on the 2-nd operation). The total number of operations would be 6 + 6 + 23 + 20 = 55. However, to make "ELLY" an anagram of "KRIS" it would be better to change it to "IRSK" with only 29 operations.  You are given the strings A and B. Find the minimal number of operations needed to transform A into some other string X, such that X is an anagram of B.

 

输入

There will be multiple test cases. For each test case:
There is two strings A and B in one line. |A| = |B| ≤ 50. A and B will contain only uppercase letters from the English alphabet (‘A‘-‘Z‘).

 

输出

For each test case, output the minimal number of operations.

样例输入

ABCA BACA
ELLY KRIS
AAAA ZZZZ

样例输出

0
29
100

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;

int main()
{
    char str1[55];
    char str2[55];
    while(scanf("%s",str1)!=EOF)
    {
        scanf("%s",str2);
        int len =strlen(str1);
        int a[26]= {0};
        int b[26]= {0};
        for(int i=0; i<len; i++)
        {
            a[str1[i]-‘A‘]++;
            b[str2[i]-‘A‘]++;
        }
        for(int i=0; i<26; i++)
        {
            int t = min(a[i],b[i]);
            a[i]-=t;
            b[i]-=t;
        }
        long long int ans = 0;
        for(int i=0; i<26; i++)
        {
            while(a[i])
            {
                for(int j=i+1;i!=j%26; j++)
                {
                    if(b[j%26]!=0)
                    {
                        if(i<=j%26)
                        {
                            ans+=j%26-i;
                           // cout<<j%26-i<<endl;
                        }
                        else
                        {
                            ans+=26-i+j%26;
                           // cout<<26-i+j%26<<endl;
                        }
                        b[j%26]--;
                        break;
                    }

                }
                a[i]--;
            }
        }
        printf("%lld\n",ans);
    }
}

注意最优解为无序的替换