hust1010 The Minimum Length

地址：http://acm.hust.edu.cn/problem/show/1010

题目：

1010 - The Minimum Length

Time Limit: 1s Memory Limit: 128MB

Submissions: 2502 Solved: 925

DESCRIPTION

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

bcabcab
efgabcdefgabcde

SAMPLE OUTPUT

3
7
思路：kmp+最小循环节

#include <bits/stdc++.h>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-8;
const double pi=acos(-1.0);
const int K=1e6+7;
const int mod=1e9+7;

int nt[K];
char sa[K],sb[K];

void kmp_next(char *T,int *nt)
{
    nt[0]=0;
    for(int i=1,j=0,len=strlen(T);i<len;i++)
    {
        while(j&&T[j]!=T[i])j=nt[j-1];
        if(T[j]==T[i])j++;
        nt[i]=j;
    }
}
int kmp(char *S,char *T,int *nt)
{
    int ans=0;
    kmp_next(T,nt);
    int ls=strlen(S),lt=strlen(T);
    for(int i=0,j=0;i<ls;i++)
    {
        while(j&&S[i]!=T[j])j=nt[j-1];
        if(S[i]==T[j])j++;
        if(j==lt)
             ans++,j=0;
    }
    return ans;
}
int main(void)
{
    int t;
    while(scanf("%s",sa)==1)
    {
        kmp_next(sa,nt);
        int len=strlen(sa);
        int ans=len-nt[len-1];
        printf("%d\n",ans);
    }
    return 0;
}