Theme Section 【KMP】

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

题目大意：给出一串主串，让求主串中EAEBA形式的E串的最大长度，必须要开头E串结尾E串，AB串长度任意，可以是0；

思路：既然要求三个相同的子串，而且有两个还必须在开头和结尾，那就求Next数组，Next数组存的是前后缀相同的长度，所以只需要找【2*i，L-Next[i]】之间相同的字串，即Next[j]==i;

具体看代码：

#include<stdio.h>
#include<string.h>
#include<string>
#include<algorithm>
#include<math.h>
#include<map>
#include<queue>
#include<vector>
#include<stack>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int N=1000005;

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

void get_Next()
{
    int i=0,j=-1;
    Next[0]=-1;
    while(i<l)
    {
        if(j==-1||s[i]==s[j])
            Next[++i]=++j;
        else
            j=Next[j];
    }
}

int KMP()
{
    int i,j;
    for(i=Next[l]; i; i=Next[i])
        for(j=2*i; j<=l-i; j++)
            if(Next[j]==i)
                return i;
    return 0;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%s",s);
        int ans=0,k,i,j;
        l=strlen(s);
        get_Next();
//        for(i=1; i<=l; i++)
//            printf("%d ",Next[i]);
//        printf("\n");
        printf("%d\n",KMP());
    }
    return 0;
}