“浪潮杯”第九届山东省ACM大学生程序设计竞赛重现赛——A

链接： https://www.nowcoder.com/acm/contest/123/A
来源：牛客网
Anagram

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 65536K，其他语言131072K
64bit IO Format: %lld

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.

输入描述:

There will be multiple test cases. For each testcase:

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’).

输出描述:

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

示例1

输入

ABCA BACA
ELLY KRIS
AAAA ZZZZ

输出

0
29
100

题目给出两个字符串A、B，这两个字符串都由'A'-'Z'构成，字符串X是字符串B任意颠倒字符顺序构成的，求由字符串A转变到字符串X的最少操作数，如A->B，1步 ；B->A，25步（可以看成Z、A相连成环）

思路：先将字符串A、B分别排序，如果a[i]<=b[j]依次寻找距离字符a[i]最近的字符b[j],如果a[i]>b[j],则将剩下的配对（这里不需要考虑配对的次序，可以自己举例）

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
char a[50],b[50];
int len,ans,num;
int judge(char a,char b)
{
	int sum=0;
	if(a<=b)sum+=b-a;
	else sum+=26-(a-b);
	return sum;
}
char cmp(char x,char y)
{
	return x<y;
}
int main()
{
	while(cin>>a>>b)
	{
		num=0;
		len=strlen(a);
		sort(a,a+len,cmp);
		sort(b,b+len,cmp);
		for(int i=0;i<len;i++)
		{
		    for(int j=0;j<len;j++)
		  { 
		   if(b[j]!=0)
		   {
		   	    if(b[j]>=a[i])
		   	    {
		   	    num+=judge(a[i],b[j]);
		   	     b[j]=0;//标记b[j]是否配对成功
		   	     a[i]=0;//标记a[i]是否配对成功
		   	     break;		
			    }
			}
		   }
		  }
		  for(int i=0;i<len;i++)
		{
			if(a[i]!=0)
		     {
				for(int j=0;j<len;j++)
		        { 
		         if(b[j]!=0)//剩下的必为a[i]>b[j]
		         {
		   	    num+=judge(a[i],b[j]);
		   	     b[j]=0;
		   	     break;	
			  }
		        }
		    }
		 }
		cout<<num<<endl;
		}
	return 0;
}