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
Input
5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba
Output
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：数据量这么大，就不能单纯的找回文串了。马拉车算法就是求最长回文子串。
这道题的解题思路为：
①同时从前后查找匹配的字符串，直到字符不匹配。
②剩下的字符串按照马拉车算法求最长回文前缀和最长回文后缀。比较得出最长的。
③如果步骤①没有找到匹配的字符，就把原字符串按照马拉车算法求得最长回文前缀和最长回文后缀。（牵扯到字符下标，因此细节有点多。）
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=1e6+100;
int p[maxx<<1];
string s,t1,t2,s1,s2,ss,_max;
int n;
inline string Manacher(string t)
{
string tt="&#";
for(int i=0;i<t.length();i++)
{
tt+=t[i];
tt+='#';
}
memset(p,0,tt.length());
int mx=0,id=0;
for(int i=0;i<tt.length();i++)
{
p[i]=mx>i?min(p[id*2-i],mx-i):1;
while(tt[i+p[i]]==tt[i-p[i]]) p[i]++;
if(i+p[i]>mx)
{
mx=i+p[i];
id=i;
}
}
//下面的操作就是求最长回文前缀和最长回文后缀
int len=0;
for(int i=2;i<tt.length();i++) if(i-(p[i]-2)==2) len=max(len,p[i]-1);
string z1=t.substr(0,len);
len=0;
for(int i=2;i<tt.length();i++) if(i+(p[i]-2)==tt.length()-2) len=max(len,p[i]-1);
string z2=t.substr(t.length()-len);
if(z1.length()>z2.length()) return z1;
else return z2;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
cin>>s;
int l=0,r=s.length()-1,L,R;
while(l<r&&s[l]==s[r]) l++,r--;
L=l,R=r;
if(l>r) swap(l,r),l++;
else if(l<r) r++;
t1=s.substr(0,l);
t2=s.substr(r);
if(t1.length()==0||t2.length()==0) _max=Manacher(s);
else if(L<R)
{
ss=s.substr(L,R-L+1);
_max=Manacher(ss);
_max=t1+_max+t2;
}
else _max=t1+t2;
cout<<_max<<endl;
}
return 0;
}
努力加油a啊，(o)/~
