（KMP灵活运用 利用Next数组 ）Theme Section -- hdu -- 4763

http://acm.hdu.edu.cn/showproblem.php?pid=4763

 

Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2129    Accepted Submission(s): 997

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

5

xy

abc

aaa

aaaaba

aaxoaaaaa

 

 

Sample Output

0

0

1

1

2

 

就是找到一个 类似 EAEBE 这样的结构， 其中找到 E 最长为多少

 

 

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>

using namespace std;
#define INF 0x3f3f3f3f
#define N 1000007

char s[N];
int Next[N];

void FindNext()
{
    int i=0, j=-1;
    int len = strlen(s);

    Next[0] = -1;

    while(i<len)
    {
        if(j==-1 || s[i]==s[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}

bool KMP(int Start, int End) 
{
    int i=0, j=Start;
    /// i 是子串的开始， j 是母串的开始
    /// 0~Start-1 是子串， Start~End-1 是母串，在母串中查找是否含有子串
    
    while(i<Start && j<End)
    {
        while(i==-1 && (s[i]==s[j] && i<Start))
            i++, j++;

        if(j==Start)
            return true;
        i = Next[i];
    }
    return false;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%s", s);

        FindNext();

        int len = strlen(s);

        int j = len;

        while(Next[j]>0)
        {
            /// next[j] 是最大前缀，就是母串的开始位置
            /// 因为前缀也是后缀， 所以用总长度减后缀就是母串结束的位置
            if(KMP(Next[j], len-Next[j])==true)
                break;
            j = Next[j];
        }

        printf("%d\n", Next[j]);
    }
   return 0;
}