Codeforces Round #277 (Div. 2)C.Palindrome Transformation 贪心

C. Palindrome Transformation

 

 

Nam is playing with a string on his computer. The string consists of n lowercase English letters. It is meaningless, so Nam decided to make the string more beautiful, that is to make it be a palindrome by using 4 arrow keys: left, right, up, down.

There is a cursor pointing at some symbol of the string. Suppose that cursor is at position i (1 ≤ i ≤ n, the string uses 1-based indexing) now. Left and right arrow keys are used to move cursor around the string. The string is cyclic, that means that when Nam presses left arrow key, the cursor will move to position i - 1 if i > 1 or to the end of the string (i. e. position n) otherwise. The same holds when he presses the right arrow key (if i = n, the cursor appears at the beginning of the string).

When Nam presses up arrow key, the letter which the text cursor is pointing to will change to the next letter in English alphabet (assuming that alphabet is also cyclic, i. e. after 'z' follows 'a'). The same holds when he presses the down arrow key.

Initially, the text cursor is at position p.

Because Nam has a lot homework to do, he wants to complete this as fast as possible. Can you help him by calculating the minimum number of arrow keys presses to make the string to be a palindrome?

Input

The first line contains two space-separated integers n (1 ≤ n ≤ 105) and p (1 ≤ p ≤ n), the length of Nam's string and the initial position of the text cursor.

The next line contains n lowercase characters of Nam's string.

Output

Print the minimum number of presses needed to change string into a palindrome.

Sample test(s)

input

8 3
aeabcaez

output

6

Note

A string is a palindrome if it reads the same forward or reversed.

In the sample test, initial Nam's string is:  (cursor position is shown bold).

In optimal solution, Nam may do 6 following steps:

The result, , is now a palindrome.

题意：给你一个长度n的字符串和光标的起始位置，

再给出以下4种操作 光标左移 光标右移 字符上换 字符下换。 问将给出的字符换成回文串的最小花费是多少。

题解：可以先预处理出每个地方的光标上下变换次数，再贪心求步数最小

我们当然贪心在一半边内移动

///
#include<bits/stdc++.h>
using namespace std ;
typedef long long ll;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
#define meminf(a) memset(a,127,sizeof(a));
 
inline ll read()
{
    ll x=,f=;char ch=getchar();
    while(ch<''||ch>''){
        if(ch=='-')f=-;ch=getchar();
    }
    while(ch>=''&&ch<=''){
        x=x*+ch-'';ch=getchar();
    }return x*f;
}
//****************************************
#define maxn 100000+50
#define inf 1000000007
 
int main(){
int n,k;
char a[maxn];
int G[maxn];
  bool bo=;
   scanf("%d%d",&n,&k);getchar();
   for(int i=;i<=n;i++){
      scanf("%c",&a[i]);
   }
   int l=,r=n,pos,next[maxn];
   if(k<=n/)pos=;
   else pos=;
   int kk=;G[]=-inf;
   while(l<=r){
     if(a[l]!=a[r]){
          if(pos==){
            G[++kk]=r;
            if(r==k)bo=;
            next[r]=l;
          }
          else{
              if(l==k)bo=;G[++kk]=l;next[l]=r;
          }
     }
 
     l++,r--;
   }
   if(!bo){
      G[++kk]=k;
   }
   sort(G+,G+kk+);
   int ans=;
   int last;//cout<<kk<<endl;
   int fa=lower_bound(G+,G+kk+,k)-G;
   if(abs(G[]-G[fa])>=abs(G[kk]-G[fa])){
        last=G[fa];//cout<<last<<" "<<G[fa]<<endl;
     for(int i=fa+;i<=kk;i++){
        ans+=min(-abs(a[G[i]]-a[next[G[i]]]),abs(a[G[i]]-a[next[G[i]]]));;
        ans+=abs(G[i]-last);last=G[i];
     }last=G[kk];//cout<<ans<<endl;
     for(int i=fa-;i>=;i--){
        ans+=min(-abs(a[G[i]]-a[next[G[i]]]),abs(a[G[i]]-a[next[G[i]]]));
       ans+=abs(G[i]-last);last=G[i];
     }//cout<<ans<<endl;
   }
   else {
    last=G[fa];
     for(int i=fa-;i>=;i--){
        ans+=min(-abs(a[G[i]]-a[next[G[i]]]),abs(a[G[i]]-a[next[G[i]]]));
        ans+=abs(G[i]-last);;last=G[i];
     } last=G[];
     for(int i=fa+;i<=kk;i++){
        ans+=min(-abs(a[G[i]]-a[next[G[i]]]),abs(a[G[i]]-a[next[G[i]]]));
        ans+=abs(G[i]-last);;last=G[i];
     }
   }if(bo)ans+=min(-abs(a[k]-a[next[k]]),abs(a[next[k]]-a[k]));
   cout<<ans<<endl;
 
  return ;
}

代码