Palindrome POJ - 3974 （字符串hash+二分） - Java

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Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?"

A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not.

The students recognized that this is a classical problem but couldn't come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said "Okay, I've a better algorithm" and before he starts to explain his idea he stopped for a moment and then said "Well, I've an even better algorithm!".

If you think you know Andy's final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.

Input

Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity).

Output

For each test case in the input print the test case number and the length of the largest palindrome.

Sample Input

abcbabcbabcba

abacacbaaaab

END

Sample Output

Case 1: 13

Case 2: 6

题意：给一个字符串，找出其中最长回文子串长度
思路：二分枚举其长度，然后枚举每个位置每个位置，利用字符串hash值比较中间位置前后是否一致，需要分奇偶进行二分，因为奇数或者偶数的时候判断前后时候一致的区间不同

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<algorithm>

 using namespace std;

 const int maxn = 1e6+;

 unsigned long long f1[maxn],f2[maxn],p[maxn];

 int even[maxn>>],odd[maxn>>];

 char word[maxn];

 int n;

 unsigned long long Find_l(int l,int r)

 {

     return f2[n-l+]-f2[n-r]*p[r-l+];//逆序hash值

 }

  unsigned long long Find_r(int l,int r)

 {

     return f1[r]-f1[l-]*p[r-l+];//正序hash值

 }

 int check(int x)

 {

     int len = x;

     x /= ;

     if(len & )

     {

         for(int i=x; i<=n-x-; i++)

             if(Find_l(i-x+,i)==Find_r(i+,i+x+))

                 return len;

     }

     else

     {

         for(int i=x; i<=n-x; i++)

             if(Find_l(i-x+,i)==Find_r(i+,x+i))

                 return len;

     }

     return ;

 }

 int serch(int num[],int l,int r)

 {

     int maxx = ;

     while(l <= r)

     {

         int mid = (l+r) >> ;

         int val = check(num[mid]);

         if(val)

         {

             l = mid + ;

             maxx = max(val,maxx);

         }

         else

             r = mid - ;

     }

     return maxx;

 }

 int main()

 {

     int tot1 = ,tot2 = ;

     p[] = ;

     for(int i=; i<=; i++)

     {

         p[i] = p[i-]*;

         if(i&)

             odd[++tot1] = i;

         else

             even[++tot2] = i;

     }

     int cas = ;

     while(~scanf("%s",word+) && word[] != 'E')

     {

         f1[] = f2[] = ;

         n = strlen(word+);

         for(int i=; i<=n; i++)

         {

             f1[i] = f1[i-]* + word[i]-'a'+;

             f2[i] = f2[i-]* + word[n-i+]-'a'+;

         }

         int ans1 = serch(odd,,n/+);

         int ans2 = serch(even,,n/+);

         printf("Case %d: %d\n",++cas,max(ans1,ans2));

     }

 }