C. Palindrome Again !!
Given string with N characters, your task is to transform it to a palindrome string. It's not as easy as you may think because there is a cost for this transformation!!
First you have to start from character at given position P. From your position you always have 2 options:
- You can move one step to the right or to the left, the cost of each movement is 1. Assume that the string is cyclic, this means if you move one step to the left you will be at position P-1 if P > 1 or at the last character if P = 1, and if you move one step to the right you will be at position P+1 if P < N or at first character if P = N.
- You can change the letter at your current position by replacing it with the next or previous one in the English alphabet (assume that the alphabet is also cyclic so ‘a’ is after ‘z’). The cost of each replacement is also 1.
You should repeat that until the transformation is finished and the string is palindrome. What is the minimum cost to do that?
The first line contains the number of test cases T ( 1 ≤ T ≤ 100 ). Each test case contains 2 lines, the first line contains two integers ( 1 ≤ N ≤ 100,000) the length of string and ( 1 ≤ P ≤ N ) the initial position. While the second line contains a string with exactly N alphabetical characters.
For each test case output one line contains the minimum cost that is needed to change the string into a palindrome one.
1
8 3
aeabdaey
8
start with P = 3 ae(a)bdaey, move right => aea(b)daey, change to next => aea(c)daey, change to next => aea(d)deay, move left => ae(a)ddeay, move left => a(e)addeay, move left => (a)eaddeay, change to previous => (z)eaddeay, change to previous => (y)eaddeay. This costs 8 (4 movements and 4 replacements)
題目連結：http://codeforces.com/gym/100952/problem/C
題目大意：給出字串a,將該字串變成迴文串！
分析：
字串向左邊或右邊移動一步（0往前移一格為n-1看成環），花費為1當前字母變為相鄰字母，例如a -> b 或 a -> z， 花費為1
模擬此過程操作即可，程式碼給出了詳細註釋，一看就懂了！
下面給出AC程式碼：
1 #include <bits/stdc++.h> 2 using namespace std; 3 inline int read() 4 { 5 int x=0,f=1; 6 char ch=getchar(); 7 while(ch<'0'||ch>'9') 8 { 9 if(ch=='-') 10 f=-1; 11 ch=getchar(); 12 } 13 while(ch>='0'&&ch<='9') 14 { 15 x=x*10+ch-'0'; 16 ch=getchar(); 17 } 18 return x*f; 19 } 20 inline void write(int x) 21 { 22 if(x<0) 23 { 24 putchar('-'); 25 x=-x; 26 } 27 if(x>9) 28 write(x/10); 29 putchar(x%10+'0'); 30 } 31 int main() 32 { 33 int t; 34 t=read(); 35 while(t--) 36 { 37 int len,p; 38 len=read(); 39 p=read(); 40 p--;//從零下標開始計數 41 string s; 42 cin>>s; 43 //移動距離 44 int minn=1e+8; 45 int maxn=-1e+8; 46 int ans=0,flag=0; 47 for(int i=0,j=len-1;i<len/2;i++,j--) 48 { 49 if(s[i]!=s[j]) 50 { 51 int a=min(s[i],s[j])-'a'; 52 int b=max(s[i],s[j])-'a'; 53 ans+=min(b-a,a+26-b);//變換需要移動的最短距離 54 minn=min(minn,i);//滿足條件最近的下標值 55 maxn=max(maxn,i);//滿足條件最遠的下標值 56 flag=1; 57 } 58 } 59 if(!flag)//如果序列已經是迴文序列 60 { 61 printf("0\n"); 62 continue; 63 } 64 if(len%2==1&&p==len/2)//奇數長度的迴文序列,查詢下標剛好是其中點時 65 { 66 ans+=abs(p-minn);//移動距離為其長度的一半 67 printf("%d\n",ans); 68 continue; 69 } 70 if(p>=len/2) 71 p=len-p-1;//這是一個迴文序列，是對稱的，所以我們採取序列號從小往大的排列方式 72 int c=abs(minn-p); 73 int d=abs(maxn-p); 74 ans+=min(c,d);//在變換過程中會忽略那些對稱的點，所以這步是要加上那些忽略點的距離 75 ans+=(maxn-minn);//起始變換點與變換終點的距離，也就是移動距離 76 printf("%d\n",ans); 77 } 78 return 0; 79 }