杭电4763 Theme Section（KMP）

Theme Section

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1380    Accepted Submission(s): 697

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

 

Source

2013 ACM/ICPC Asia Regional Changchun Online 

/*ben lai hai xiang da fa gan kai ne ,jie guo shu ru fa chu wen ti le,wu yu .....shen qi de KMP  Time:2014-11-22 15:08*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAX=1000000+2;int next[MAX];char s[MAX];void Get_next(){int i=0,j=-1;next[0]=-1;//将首位置初始化为-1 while(s[i]){if(j==-1||s[i]==s[j]){i++;j++;next[i]=j; /*注意：此题不能用优化后的算法，因为是拓展next求自身前后匹配, 优化算法加的条件语句 if(s[i]==s[j])next[i]=next[j],因为值越小，跳跃越大 */ }else{j=next[j];}}}int kmp(){Get_next();int len=strlen(s);/*总长度的next值即为自身匹配的最大值*/int i,j;for(i=next[len];i;i=next[i]){//不存在的话寻找到0跳出 for(j=i>>1;j<=len-i;j++){//len应是<= if(next[j]==i)//中间如果能找到next的值为i的，则证明，存在第三个 return i;}}}int main(){int n;scanf("%d",&n);while(n--){ memset(s,0,sizeof(s));scanf("%s",s);printf("%d\n",kmp());}return 0;}