【思维】hdu 6103 Kirinriki

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本文介绍了一个关于字符串的问题，即如何寻找两个非重叠子串，使它们的定义距离小于等于给定值，并讨论了相应的解决算法。通过从中心向两端枚举的方法，实现了高效的求解。

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Kirinriki

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1792 Accepted Submission(s): 735

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

  

   
    
     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

题意：输入m和一个串，要在串中找到两个不会重叠长度相等并且长度尽量长的字串，使得定义的距离小于等于m；
思路：从中心向两端枚举，学弟画了一张图(抱大腿啊)：

找每条斜线上的距离的和小于等于m的长度最长是多少，最后去最长即可；

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

string s1,s2;
int s[5050],m;
int sum[5050];

int get_len(int n)  //尺取
{
    sum[0]=0;
    for(int i=1; i<=n; i++)
        sum[i]=sum[i-1]+s[i];

    int maxx=0,l=1,r=1;
    while(l<=r&&l<=n&&r<=n)
    {
        if(sum[r]-sum[l-1]<=m)
        {
            maxx=max(maxx,r-l+1);
            r++;
        }
        else
        {
            l++;
            if(l>r) //没加这句，wa晕
                r=l;
        }
    }
    return maxx;
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(s,0,sizeof(s));

        int max_len=0;
        scanf("%d",&m);
        cin>>s1;
        s2=s1;
        reverse(s2.begin(),s2.end());

        for(int i=0; i<(int)s1.size()-1; i++)
        {
            int len=(s2.size()-i)/2;
            if(max_len>=len) break;

            for(int j=0; j<len; j++)
                s[j+1]=abs(s1[i+j]-s2[j]);

            max_len=max(max_len,get_len(len));
        }

        for(int i=0; i<(int)s2.size()-1; i++)
        {
            int len=(s1.size()-i)/2;
            if(max_len>=len) break;

            for(int j=0; j<len; j++)
                s[j+1]=abs(s2[i+j]-s1[j]);

            max_len=max(max_len,get_len(len));
        }

        printf("%d\n",max_len);
    }
    return 0;
}