hust 1010 The Minimum Length (KMP)

1010 - The Minimum Length

时间限制：1秒 内存限制：128兆

3105 次提交 1148 次通过

题目描述

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.

输入

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.

输出

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

样例输入

bcabcabefgabcdefgabcde

样例输出

37

分析：由于这个串是全部由一个子串连成的，所以用主串的长度减去最后一个的next值就是子串的长度。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define MAX_N 1000005using namespace std;char p[MAX_N];int ne[MAX_N];void makeNext(const char p[],int ne[]){    int len=strlen(p);    ne[0]=0;    for(int i=1,k=0;i<len;i++)    {        while(k>0&&p[i]!=p[k]) k=ne[k-1];        if(p[i]==p[k]) k++;        ne[i]=k;    }}int main(){    while(~scanf("%s",p))    {        makeNext(p,ne);        int len=strlen(p);        printf("%d\n",len-ne[len-1]);    }    return 0;}