hdu 6103 Kirinriki (枚举对称中心+双指针)

Problem Description

We define the distance of two strings A and B with same length n is
disA,B=∑(i=0  n−1)  |A[i]−B[n−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

 

给你一个m值,一个字符串,让你在字符串中找两个不重叠的长度相等的字串,两个字符串首对应尾逐项向中心的acsii码差的绝对值之和不能大于m

问这两个相等的字符串最长有多长

 

我们枚举这两个字串的对称中心,对于每个对称中心我们利用双指针进行处理一下就好了

代码如下:

#include <bits/stdc++.h>

using namespace std;
const int maxn = 5000+50;
char s[maxn];
int m,len;
int ans;
void work (int x,int y)
{
    int dis=0,l=0,r=0;//左边的串s[x-l]~s[x-r] 右边的串s[y+l]~s[y+r]
    while (y+r<len&&x-r>=0){
        if (dis+abs(s[x-r]-s[y+r])<=m){
            dis+=abs(s[x-r]-s[y+r]);
            r++;
            ans = max(ans,r-l);
        }
        else{
                dis-=abs(s[x-l]-s[y+l]);
                l++;
        }
    }
}
int main()
{
    //freopen("de.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while (T--){
        scanf("%d",&m);
        scanf("%s",s);
        len = strlen(s);
        ans = 0;
        for (int i=0;i<len;++i){
            work(i-1,i+1);
            work(i,i+1);
        }
        printf("%d
",ans);
    }
    return 0;
}