Bis longest palindromic sequence

# The meaning of problems

Given a string s, s seeking longest string palindromic double t, can be divided into two parts t x, y and x, y are palindromic sequence

# Explanations

And seeking l l r array [i] on the basis manacher and r [i] point to the left represent the axis i of the extended string and the right endpoint palindrome

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 const int maxn=200010;
 8 char a[maxn],s[maxn<<1];
 9 int l[maxn<<1],r[maxn<<1];
10 int n,hw[maxn],ans;
11 inline void manacher() {
12     int maxright=0,mid;
13     for(int i=1; i<n; ++i) {
14         if(i<maxright)
15             hw[i]=min(hw[(mid<<1)-i],hw[mid]+mid-i);
16         else
17             hw[i]=1;
18         while(s[i+hw[i]]==s[i-hw[i]])
19             ++hw[i];
20         if(hw[i]+i>maxright) {
21             maxright=hw[i]+i;
22             mid=i;
23         }
24     }
25 }
26 
27 void change() {
28     s[0]=s[1]='#';
29     for(int i=0; i<n; ++i) {
30         s[i*2+2]=a[i];
31         s[i*2+3]='#';
32     }
33     n=n*2+2;
34     s[n]=0;
35 }
36 
37 inline int maxx(int a,int b) {
38     return a>b?a:b;
39 }
40 
41 int main() {
42     scanf("%s",a);
43     n=strlen(a);
44     change();
45     manacher();
46     int now=0;
47     for(int i=0; i<n; ++i)
48         while(now<=i+hw[i]-1)
49             l[now++]=i;
50     now=n;
51     for(int i=n-1; i>=0; --i) {
52         while(now>=i-hw[i]+1)
53             r[now--]=i;
54     }
55     int ans=0;
56     for(int i=0; i<n; ++i)
57         ans=maxx(ans,r[i]-l[i]);
58     printf("%d",ans);
59 }