hust 1010 KMP算法求最小循环节

1010 -- The Minimum Length

Time Limit: 1S Memory Limit: 128MB

Submissions: 534 Solved: 186

DESCTIPTION

There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B. Then, give you the string B, can you tell me the length of the shortest possible string A. For example, A="abcdefg". I got abcdefgabcdefgabcdefgabcdefg.... Then I cut the red part: efgabcdefgabcde as string B. From B, you should find out the shortest A.

INPUT

Multiply Test Cases. For each line there is a string B which contains only lowercase and uppercase charactors. The length of B is no more than 1,000,000.

OUTPUT

For each line, output an integer, as described above.

SAMPLE INPUT

bcabcabefgabcdefgabcde

SAMPLE OUTPUT

37

HINT

SOURCE

这个题目的意思是说给出一a字符串，多次复制，取出其中一部分为b,求出其中最小长度的a是多少，实际上这个题目的意思是在求一个字符串中的最小循环节，求出next数组，取next[t],也就是整个字符串中前缀和后缀的最大匹配值，列如bcabcab中next[7]=4,其中bcab,和bcab,其中b被重复利用了两次，所以可以求出其中的最小循环节就是7-4=3,也就是cab了，今天早上吃面条时吃玲玲小朋友讲了kmp,好不容易终于会了，实在不容易，这个面条吃的值，吃的值！！！！

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int t;char s[1000005];int next[1000005];void getNext(){     int j=-1;     int i=0;     next[0]=-1;     t=strlen(s);     while(i<t)     {       if((j==-1)||(s[i]==s[j]))       {           i++;           j++;           next[i]=j;       }       else       {           j=next[j];       }     }}int main(){    int i,j,k;    while(scanf("%s",s)!=EOF)    {     t=strlen(s);     getNext();     printf("%d\n",t-next[t]);    }    return 0;}