poj3974 Palindrome【回文】【Hash】

Palindrome

Time Limit: 15000MS		Memory Limit: 65536K
Total Submissions: 13157		Accepted: 5028

Description

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

Source

Seventh ACM Egyptian National Programming Contest

题意：求一个字符串的最长回文子串

思路：回文串其实就是以一个节点为中间，两端的字符串是相同的。之前的比较字符串相同的Hash函数是以从左到右的顺序，那么这个就再存一个从右到左的字符串的Hash值。对于每一个字符，二分左半子串的长度，分回文串的长度是奇还是偶两种情况。

 1 #include <iostream>
 2 #include <set>
 3 #include <cmath>
 4 #include <stdio.h>
 5 #include <cstring>
 6 #include <algorithm>
 7 #include <map>
 8 using namespace std;
 9 typedef long long LL;
10 #define inf 0x7f7f7f7f
11 
12 const int maxn = 1e6 + 5;
13 char s[maxn];
14 unsigned long long H[maxn], p[maxn], H_rev[maxn];
15 
16 unsigned long long getH(int i, int j)
17 {
18     return H[j] - H[i - 1] * p[j - i + 1];
19 }
20 
21 unsigned long long getHrev(int i, int j)
22 {
23     return H_rev[i] - H_rev[j + 1] * p[j - i + 1];
24 }
25 
26 int main()
27 {
28     int cas = 1;
29     p[0] = 1;
30     for(int i = 1; i < maxn; i++){
31         p[i] = p[i - 1] * 131;
32     }
33     while(scanf("%s", s + 1)){
34         if(strcmp(s + 1, "END") == 0){
35             break;
36         }
37         int n = strlen(s + 1);
38         H[0] = 0;
39         H_rev[n + 1] = 0;
40         for(int i = 1; i <= n; i++){
41             H[i] = H[i - 1] * 131 + (s[i] - 'a' + 1);
42         }
43         for(int i = n; i >= 1; i--){
44             H_rev[i] = H_rev[i + 1] * 131 + (s[i] - 'a' + 1);
45         }
46 
47         int ans = -1;
48         for(int i = 1; i <= n; i++){
49             int ped = min(i - 1, n - i), pst = 1;
50             while(pst < ped){
51                 int pmid = (pst + ped + 1) / 2;
52                 //cout<<pmid<<endl;
53                 if(getH(i - pmid, i - 1) == getHrev(i + 1, i + pmid)){
54                     //
55                     pst = pmid;
56                 }
57                 else{
58                     ped = pmid - 1;
59                 }
60             }
61             //cout<<i<<" "<<pst<<endl;
62             ans = max(ans, 2 * pst + 1);
63             int qed = min(i - 1, n + 1 - i), qst = 1;
64             while(qst < qed){
65                 int qmid = (qst + qed + 1) / 2;
66                 if(getH(i - qmid, i - 1) == getHrev(i, i + qmid - 1)){
67 
68                     qst = qmid;
69                 }
70                 else{
71                     qed = qmid - 1;
72                 }
73             }
74             ans = max(ans, 2 * qst);
75 
76         }
77 
78         printf("Case %d: %d\n", cas++, ans);
79     }
80 
81 }