hdu6103Kirinriki（第六场尺取法取区间最大）

Kirinriki

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1822    Accepted Submission(s): 749

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

求最大字串，使得交叉相减绝对值和最大且不超过m的字串长度

处理区间最大问题用尺取法，从a开始枚举，如果枚举的res值超过m，则退回到开头，下次该区间跳过枚举，枚举到最后一个字符时，枚举起始地址+1。

下面举例说明：

1
2
abcd
####0-1=1
####0-2=3
****0-2=1
l ---->0
^^^^0-2=-1
####0-2=1
####0-3=4
****0-3=1
l ---->0
^^^^0-3=-2
####0-3=1
####1-2=2
2

带#号的为第一次枚举，*号为超出了m，去掉当前枚举值，^为继续后退，得到新的起始值，继续枚举

最后直到枚举到1-2结束，输出ans=2

#include <stdio.h>#include <cstring>#include <algorithm>#include <iostream>using namespace std;const int maxn=20005;char s[maxn];int m,len;int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&m);        scanf("%s",s);        len=strlen(s);        int ans=0;        for(int i=0;i<len;i++)        {            int t=0,res=0,l=0;            for(int j=0;j<(i+1)/2;j++)            {                res+=abs(s[j]-s[i-j]);                //printf("####%d-%d=%d\n",j,i-j,ans);                if(res<=m)                {                    t++;                    ans=max(t,ans);                }                else                {                    res-=abs(s[j]-s[i-j]);                    //printf("****%d-%d=%d\n",j,i-j,ans);                    res-=abs(s[l]-s[i-l]);                    //printf("l ---->%d\n",l);                    //printf("^^^^%d-%d=%d\n",j,i-j,ans);                    l++,t--,j--;                }            }        }        reverse(s,s+len);        //cout<<s<<endl;        for(int i=0;i<len;i++)        {            int t=0,res=0,l=0;            for(int j=0;j<(i+1)/2;j++)            {                res+=abs(s[j]-s[i-j]);                if(res<=m)                {                    t++;                    ans=max(t,ans);                }                else                {                    res-=abs(s[j]-s[i-j]);                    res-=abs(s[l]-s[i-l]);                    l++,t--,j--;                }            }        }        printf("%d\n",ans);    }}