HDU 4763 Theme Section【KMP的next数组练习】

Theme Section

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4030 Accepted Submission(s): 1906

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

提议概括：

在保证有相同前后缀的前提下，在字符串中间找到与前后缀相同的子串，并输出前后缀和中间相同长度的最大值。

解题分析：

先用KMP的方法建立一个next数组，用来记录前后缀相同的情况，然后通过对next数组的查找判断中间相同的部分。

AC代码：

#include<stdio.h>#include<string.h>#define N 1000010char str[N];int next[N], len;int get_next(void){    int i = 1, j = 0;    while(i < len){        if(j == 0 && str[i] != str[j]){            next[i] = 0;            i++;        }        else if(j > 0 && str[i] != str[j])            j = next[j-1];        else{            next[i] = j+1;            i++;            j++;        }    }    return next[len-1];}int main(){    int i, k, T;    scanf("%d", &T);    while(T--){        scanf("%s", str);        len = strlen(str);        k = get_next();        if(!k)            printf("0\n");        else{            for(i = len-1; next[i] > 0; i = next[i-1]){                if(next[i]*3 <= len && next[i]*2 <= i+1 || k == len-1 && next[i]*3 <= len){                    printf("%d\n", next[i]);                    break;                }            }        }    }    return 0;}