HDU6103-Kirinriki

Kirinriki

                                                                     Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                             Total Submission(s): 1691    Accepted Submission(s): 689

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

15abcdefedcb

 

Sample Output

5
Hint
[0, 4] abcde[5, 9] fedcbThe 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：给你一个字符串，让你找出两个尽量长的不重叠的子串，并且字串距离不大于m

解题思路：尺取法，用初始串做出一个反转串，然后先对初始串进行枚举，每次枚举先让两个指针指向反转串的第一个字符，先移动尾指针直至和大于m，然后移动一次头指针，再去移动尾指针直至和又大于m，不断循环做这个操作。最后再对反转串进行枚举，操作和初始串一样

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>#include <string>#include <cmath>#include <algorithm>#include <vector>#include <bitset>#include <stack>#include <queue>#include <unordered_map>#include <functional>using namespace std;char s1[5009],s2[5009];int m;int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%s",&m,s1+1);        int n = strlen(s1 + 1);        for(int i=1; i<=n; i++) s2[i]=s1[n-i+1];        s2[n+1]='\0';        int ma=0;        for(int i=1; i<=n; i++)        {            int l=1,r=1,k=0;            while(1)            {                while(k+abs(s1[i+r-1]-s2[r])<=m&&r+r+i-1<=n)                {                    k+=abs(s1[i+r-1]-s2[r]);                    r++;                }                ma=max(ma,r-l);                if(r+r+i-1>n) break;                k-=abs(s1[i+l-1]-s2[l]);                l++;            }        }        for(int i=1; i<=n; i++)        {            int l=1,r=1,k=0;            while(1)            {                while(k+abs(s2[i+r-1]-s1[r])<=m&&r+r+i-1<=n)                {                    k+=abs(s2[i+r-1]-s1[r]);                    r++;                }                ma=max(ma,r-l);                if(r+r+i-1>n) break;                k-=abs(s2[i+l-1]-s1[l]);                l++;            }        }        printf("%d\n",ma);    }    return 0;}