hdu 6103 Kirinriki（尺取法）

Kirinriki

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 933    Accepted Submission(s): 377

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

Source

2017 Multi-University Training Contest - Team 6

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题意：

给出字符串s，寻找其两个长度相同且不重叠的子串，满足其每位的ascil差值之和不大于m，且长度最长。

题解：

枚举头尾，向里缩进，使用取尺法

 

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<cstring>
#include<bits/stdc++.h>

using namespace std;
int t,m,l,r,ans,len;
char ch[5005];

void solve(int x,int y) //x，y表示起始时两个初始的头尾
{
   int l=0,r=0,dis=0; //l表示头x向里缩进l位，r表示尾y向里缩进r位
   while(x+r<y-r) //两个子串不能重合
   {
       if (dis+abs(ch[x+r]-ch[y-r])<=m)
       {
           dis+=abs(ch[x+r]-ch[y-r]);
           r++;
           ans=max(ans,r-l);
       } else
       {
           dis-=abs(ch[x+l]-ch[y-l]);
           l++;
       }
   }
}

int main()
{
    scanf("%d",&t);
    for(;t>0;t--)
    {
        scanf("%d",&m);
        scanf("%s",&ch);
        len=strlen(ch);
        ans=0;
        for(int i=0;i<len;i++)  //i表示两个子串从起始位置相差i位
        {
            solve(0,len-i-1);//头为0，尾为len-i-1，向里缩进
            solve(0+i,len-1);//头为0+i，尾为len-1，向里缩进
        }
        printf("%d\n",ans);
    }
    return 0;
}