HUST 1010 The Minimum Length（KMP）

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

题解：

题意：

给你一个由串A组成的串B

串B为AAAAAAAAAA去掉一些东西

问你原来的A最短是多长

思路：

在组成B还没删数之前，可以发现B串无论从任意一个位置开始都是循环的，删了一些字符，那么自身还是存在循环的部分，那么这个问题就转化为了对B串求最小的循环节了

直接KMP搞搞，目前HUST已经炸了2个星期了现在还没好我也不知道我这份代码对不对。。。

代码：

#include<algorithm>#include<iostream>#include<cstring>#include<stdio.h>#include<math.h>#include<string>#include<stdio.h>#include<queue>#include<stack>#include<map>#include<vector>#include<deque>using namespace std;#define lson k*2#define rson k*2+1#define M (t[k].l+t[k].r)/2#define INF 1008611111#define ll long long#define eps 1e-15int Next[1000005];char T[1000005];void init(){    int i,j;    i=0,j=-1;    Next[0]=-1;    while(T[i])    {        if(j==-1||T[i]==T[j])        {            i++;            j++;            Next[i]=j;        }        else            j=Next[j];    }}int main(){    int i,j,n;    while(scanf("%s",&T)!=EOF)    {        init();        printf("%d\n",strlen(T)-Next[strlen(T)]);    }    return 0;}