HDU 6103 Kirinriki（尺取）

最新推荐文章于 2020-11-17 09:32:09 发布

Expecto__Patronum

最新推荐文章于 2020-11-17 09:32:09 发布

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本文介绍了一道关于字符串匹配的算法题目，任务是在给定的字符串中找出两个非重叠子串，使它们之间的特定距离不超过预设限制，并求解这两个子串的最大可能长度。文章提供了问题的定义、输入输出格式示例及一种基于滑动窗口技术的解决方案。

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Kirinriki

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

Source

2017 Multi-University Training Contest - Team 6

题意：给出一个字符串，从中选取两个不重复的子串，使得dis距离小于等于m，输出最长子串的长度。

题解：类似处理回文串，枚举中心点（要分奇数偶数两种情况枚举），进行尺取，向外延伸，如果超过了m，去掉中心处位置。双指针维护。同时记录最大子串长度。时间复杂度O(n^2）

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<math.h>
#define INF 5e9;
using namespace std;
char s[5005];
int get(int i,int j)
{
    if(s[i]-s[j]<0)
        return s[j]-s[i];
    else return s[i]-s[j];
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int m;
        scanf("%d",&m);
        scanf("%s",s);
        int n=strlen(s);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            int l=i,r=i;
            int ll=i,rr=i;
            int sum=0;
            while(1)
            {
                sum+=get(l,r);
                while(sum>m&&ll>=l&&rr<=r)
                {
                    sum-=get(ll,rr);
                    ll--;
                    rr++;
                }
                if(rr==i) ans=max(r-rr,ans);
                else ans=max(r-rr+1,ans);
                if(l==0||r==n-1) break;
                l--;r++;
                
            }

        }
        for(int i=0;i<n-1;i++)
        {
            int l=i,r=i+1;
            int ll=i,rr=i+1;
            int sum=0;
            while(1)
            {
                sum+=get(l,r);
                while(sum>m&&ll>=l&&rr<=r)
                {
                    sum-=get(ll,rr);
                    ll--;
                    rr++;
                }
                ans=max(r-rr+1,ans);
                if(l==0||r==n-1) break;
                l--;r++;
                
            }

        }
        cout<<ans<<endl;

    }
    return  0;
}