Problem Description
We define the distance of two strings A and B with same length n is
disA,B=∑i=0n−1|Ai−Bn−1−i|
The difference between the two characters is defined as the difference in ASCII.
You should find the maximum length of two non-overlapping substrings in given string S, and the distance between them are less then or equal to m.
 

 

Input
The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integers m : the limit distance of substring.
Then a string S follow.

Limits
T≤100
0≤m≤5000
Each character in the string is lowercase letter, 2≤|S|≤5000
∑|S|≤20000
 

 

Output
For each test case output one interge denotes the answer : the maximum length of the substring.
 

 

Sample Input
1
5
abcdefedcb
 

 

Sample Output
5
Hint
[0, 4] abcde [5, 9] fedcb The distance between them is abs('a' - 'b') + abs('b' - 'c') + abs('c' - 'd') + abs('d' - 'e') + abs('e' - 'f') = 5
 
 
启发博客：http://www.cnblogs.com/coded-ream/p/7343946.html
以下题解和题意摘自此博客

题目描述：

找两个不重叠的字符串A，B。 使得dis（A，B）<=m;dis(A,B)=∑i=0n−1|Ai−Bn−1−i|。求最长的字符串长度。

思路：

官方题解，双指针维护。简单题。枚举对称中心。

 双指针维护！！以后要搜超时就记这个！！
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<queue>
 5 #include<map>
 6 #include<vector>
 7 #include<cmath>
 8 #include<cstring>
 9 using namespace std;
10 int m,ans,len;
11 char str[5005];
12 
13 void solve(int x,int y)//x,y表示左右端点
14 {
15     int dis=0;
16     int l=0,r=0;//双指针,r记录的是两段各自的长度,l记录的是头尾各自缩进多少
17     while(x+r<y-r)
18     {
19         if(dis+abs(str[x+r]-str[y-r])<=m)
20         {
21             dis+=abs(str[x+r]-str[y-r]);
22             r++;
23             ans=max(ans,r-l);
24         }
25         else
26         {
27             dis-=abs(str[x+l]-str[y-l]);
28             l++;
29         }
30     }
31 }
32 
33 int main()
34 {
35     int T;
36     scanf("%d",&T);
37     while(T--)
38     {
39         scanf("%d",&m);
40         cin>>str;
41         ans=0;
42         len=strlen(str);
43         for(int i=1;i<len;i++)
44             solve(0,i);//相当于枚举对称轴在前半段
45         for(int i=0;i<len-1;i++)
46             solve(i,len-1);//相当于枚举对称轴在后半段
47         printf("%d\n",ans);
48     }
49     return 0;
50 }