http://www.bnuoj.com/bnuoj/problem_show.php?pid=29375

【题意】:可以对两字符串进行如下操作:

  1、可以无损耗交换相邻两个字符(可以理解成交换任意字符)

2、可以改变一个字符 x->y ,花费为 x-y 的绝对值

求花费最少,将两字符串变成一样

【题解】:

排序字符串,然后对应相减

【code】:

 #include <iostream>
#include <stdio.h>
#include <math.h>
#include <algorithm> using namespace std; char str1[],str2[]; int abs(int a)
{
return a<?-a:a;
} int main()
{
int t,cas=;
scanf("%d",&t);
while(t--)
{
int m;
scanf("%d",&m);
scanf("%s",str1);
scanf("%s",str2);
sort(str1,str1+m);
sort(str2,str2+m);
int i,ans=;
for(i=;i<m;i++)
{
ans+=abs(str1[i]-str2[i]);
}
printf("Case %d: %d\n",cas++,ans);
}
return ;
}

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