[Codeforces Global Round 7] D2. Prefix-Suffix Palindrome (Hard version) (回文串，字符串hash)

题意

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

1. The length of t does not exceed the length of s.
2. t is a palindrome.
3. 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.
   记找s的一个不重叠的前缀a和后缀b,使得a+b是回文串且尽量长

样例

input：
5
a
abcdfdcecba
abbaxyzyx
codeforces
acbba

output:
a
abcdfdcba
xyzyx
c
abba

首先前后一一匹配，比方说abccbba，找出前缀ab后缀ba，剩下的ccb找一个最长回文前缀（cc），后缀（b）选最长的

最长回文前缀问题可以用字符串哈希解决，求一遍正向哈希，再求一遍倒着的哈希，如果是回文hash值相同，这里用的是BDKRHash，它可以O(1)求子串hash值，另外hash是非完美算法，可以在O(n)检验一下是不是回文串

#include <iostream>
#include <cstdio>
using namespace std;

typedef unsigned long long LL;
#define base 33
LL p[1000100];
LL h1[1000100],h2[1000100];

inline LL gethash1(int l,int r) {
    LL tp = l?h1[l-1]:0;
    return h1[r] - tp*p[r-l+1];
}

inline LL gethash2(int l,int r) {
    LL tp = l?h2[l-1]:0;
    return h2[r] - tp*p[r-l+1];
}

bool Is_palindrome(string s) {
    for (int i = 0; i < s.size()/2; i++)
        if (s[i] != s[s.size()-1-i]) return 0;
    return 1;
}

string find(string s) {
    string ans = "";
    int len = s.size();
    if (len == 0) return ans;
    for (int i = 0; i < len; i++) {
       if (i == 0) h1[i] = s[i];
       else h1[i] = h1[i-1]*base+s[i];
//	   cout << h1[i] << " ";
    }
//	puts("");
    // 12345
    for (int i = 0; i < len; i++) {
        if (i == 0) h2[i] = s[len-i-1];
        else h2[i] = h2[i-1]*base+s[len-i-1];
    //    cout << h2[i] << " ";
    }
    // 54321
//	puts("");
    //cout << gethash1(2,2) << " " << gethash2 (2,2) << "def\n";
//	int a = 0,b = 0;
    for (int i = len-1; i >= 0; i--) {
        if (gethash1(0,i) == gethash2(len-1-i,len-1) && Is_palindrome(s.substr(0,i+1))) return s.substr(0,i+1);
    //	if (gethash1(len-1-i,len-1) == gethash2(0,i)) return s.substr(len-1-i,i+1);
    //	cout << i << "\n";
    }
    return ans;
}

void work() {
    string s;
    cin >> s;
    string comm = "";
    int i;
    for (i = 0; i < s.size()/2; i++)
      if (s[i] == s[s.size()-1-i]) comm += s[i];
      else break;
    s = s.substr(i,s.size()-i-i);
    string ss = "";
    for (int i = s.size()-1; i>=0; i--) ss += s[i];
    string s1 = find(s),s2 = find(ss);
    cout << comm  << (s1.size()>s2.size()?s1:s2);
    for (int i = comm.size()-1; i >= 0; i--)
      cout << comm[i];
    puts("");
}

int main() {
    p[0] = 1;
    for (int i = 1; i <= 1000000; i++)
       p[i] = p[i-1]*base;

    int n;
    cin >> n;
    while (n--) {
        work();
    }
    return 0;
}