（KMP 模板）Number Sequence -- Hdu -- 1711

http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16080    Accepted Submission(s): 7100

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

 

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

 

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

 

Sample Input

2

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 1 3

13 5

1 2 1 2 3 1 2 3 1 3 2 1 2

1 2 3 2 1

 

Sample Output

6

-1

 

Source

HDU 2007-Spring Programming Contest

Next里面存的是前缀和后缀的最大相似度

Next[i] 代表的是前 i 个数的最大匹配

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define N 1000007

int M[N], S[N], Next[N];

void FindNext(int Slen)
{
    int i=0, j=-1;
    Next[0] = -1;

    while(i<Slen)
    {
        if(j==-1 || S[i]==S[j])
            Next[++i] = ++j;
        else
            j = Next[j];
    }
}



int KMP(int Mlen, int Slen)
{
    int i=0, j=0;

    FindNext(Slen);

    while(i<Mlen)
    {
        while(j==-1 || (M[i]==S[j] && i<Mlen && j<Slen))
            i++, j++;
        if(j==Slen)
            return i-Slen+1;

        j = Next[j];
    }
    return -1;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, m, i;

        scanf("%d%d", &n, &m);

        for(i=0; i<n; i++)
            scanf("%d", &M[i]);
        for(i=0; i<m; i++)
            scanf("%d", &S[i]);

        printf("%d
", KMP(n, m));
    }
    return 0;
}