HDU 4763 (KMP)

Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2721    Accepted Submission(s): 1248

Problem Description

It's time for music! A lot of popular musicians are invited to join us in the music festival. Each of them will play one of their representative songs. To make the programs more interesting and challenging, the hosts are going to add some constraints to the rhythm of the songs, i.e., each song is required to have a 'theme section'. The theme section shall be played at the beginning, the middle, and the end of each song. More specifically, given a theme section E, the song will be in the format of 'EAEBE', where section A and section B could have arbitrary number of notes. Note that there are 26 types of notes, denoted by lower case letters 'a' - 'z'.

To get well prepared for the festival, the hosts want to know the maximum possible length of the theme section of each song. Can you help us?

 

Input

The integer N in the first line denotes the total number of songs in the festival. Each of the following N lines consists of one string, indicating the notes of the i-th (1 <= i <= N) song. The length of the string will not exceed 10^6.

 

Output

There will be N lines in the output, where the i-th line denotes the maximum possible length of the theme section of the i-th song.

 

Sample Input

5xyabcaaaaaaabaaaxoaaaaa

 

Sample Output

00112

 

题意：求一个串的最长的相等的前缀和后缀，这个前缀和后缀不能相交。

因为如果可以相交的话求最长前缀和后缀非常容易，所以稍微处理一下使得最长前缀和后

缀不会交就可以了，就是中间加一个都不能匹配的字符。然后就能直接next[n]求出这个

最大前缀的长度了。

#include <cstring>#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>using namespace std;#define maxn 1111111char T[maxn], P[maxn];int n, m;#define next Nextint next[maxn];void get_next (char *p) {      int m = strlen (p);    int t;      t = next[0] = -1;      int j = 0;      while (j < m) {          if (t < 0 || p[j] == p[t]) {//匹配              j++, t++;              next[j] = t;        }          else //失配              t = next[t];      } }  int main () {    int t;    cin >> t;    while (t--) {         scanf ("%s", T);        n = strlen (T);        if (n&1)            T[n>>1] = '*';        else {            for (int i = n-1; i >= n/2; i--) T[i+1] = T[i];             n++;            T[n>>1] = '*';            T[n] = '\0';        }        get_next (T);        cout << next[n] << "\n";    }    return 0;}