Codeforces Global Round 7 D1. Prefix-Suffix Palindrome (Hard version)

D2. Prefix-Suffix Palindrome (Hard version)
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
This is the hard version of the problem. The difference is the constraint on the sum of lengths of strings and the number of test cases. You can make hacks only if you solve all versions of this task.

You are given a string s, consisting of lowercase English letters. Find the longest string, t, which satisfies the following conditions:

The length of t does not exceed the length of s.
t is a palindrome.
There exists two strings a and b (possibly empty), such that t=a+b ( “+” represents concatenation), and a is prefix of s while b is suffix of s.
Input
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤105), the number of test cases. The next t lines each describe a test case.

Each test case is a non-empty string s, consisting of lowercase English letters.

It is guaranteed that the sum of lengths of strings over all test cases does not exceed 106.

Output
For each test case, print the longest string which satisfies the conditions described above. If there exists multiple possible solutions, print any of them.

Example
inputCopy
5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba
outputCopy
a
abcdfdcba
xyzyx
c
abba
Note
In the first test, the string s=“a” satisfies all conditions.

In the second test, the string “abcdfdcba” satisfies all conditions, because:

Its length is 9, which does not exceed the length of the string s, which equals 11.
It is a palindrome.
“abcdfdcba” = “abcdfdc” + “ba”, and “abcdfdc” is a prefix of s while “ba” is a suffix of s.
It can be proven that there does not exist a longer string which satisfies the conditions.

In the fourth test, the string “c” is correct, because “c” = “c” + “” and a or b can be empty. The other possible solution for this test is “s”.
思路：
先從兩邊取對稱的前後綴，之後再取餘下字符串較長的迴文前綴或後綴。

#include <bits/stdc++.h>
using namespace std;
bool ok(const string &s,int l,int r){
    while(l<=r&&s[l]==s[r]) ++l,--r;
    return l>r;
}
void solve(){
    string s;cin>>s;
    int l=0,r=s.size()-1;
    while(l<r&&s[l]==s[r]) ++l,--r;
    int r2,l2;
    for(r2=r;r2>=l;r2--) if(ok(s,l,r2)) break;
    for(l2=l;l2<=r;l2++) if(ok(s,l2,r)) break;
    cout<<s.substr(0,l)
        <<((r2-l>r-l2)?s.substr(l,r2-l+1):s.substr(l2,r-l2+1))
        <<s.substr(r+1)
        <<"\n";
}
int main()
{
    int t;cin>>t;
    while(t--)
        solve();
    return 0;
}

大佬的博客
https://www.cnblogs.com/Kanoon/p/12528997.html