HDU 6103 17多校6 Kirinriki(双指针维护)
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.
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
[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
题目描述:
找两个不重叠的字符串A,B。 使得dis(A,B)<=m;dis(A,B)=∑i=0n−1|Ai−Bn−1−i|。求最长的字符串长度。
思路:
官方题解,双指针维护。简单题。枚举对称中心。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<cmath>
#include<cstring>
using namespace std;
int m,ans,len;
char str[]; void solve(int x,int y)//x,y表示左右端点
{
int dis=;
int l=,r=;//双指针,r记录的是两段各自的长度,l记录的是头尾各自缩进多少
while(x+r<y-r)
{
if(dis+abs(str[x+r]-str[y-r])<=m)
{
dis+=abs(str[x+r]-str[y-r]);
r++;
ans=max(ans,r-l);
}
else
{
dis-=abs(str[x+l]-str[y-l]);
l++;
}
}
} int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&m);
cin>>str;
ans=;
len=strlen(str);
for(int i=;i<len;i++)
solve(,i);//相当于枚举对称轴在前半段
for(int i=;i<len-;i++)
solve(i,len-);//相当于枚举对称轴在后半段
printf("%d\n",ans);
}
return ;
}
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