hdu Number Sequence（KMP）

  

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

没什么好解释的，直接模板

代码：
#include<stdio.h>
#define M 10001
#define N 1000001
__int64 a[N],b[M];
int next[M];
int  t,n,m;
void get_next()     
{
    int i=-1,j=0;
    next[0] = -1;
    while(j<m-1)
    {
        if(i==-1 || b[i]==b[j])
        {
            i++;
            j++;
            next[j] = i;
        }
        else i=next[i];
    }
}

int kmp()
{
    int i=0,j=0;
    while(i<n && j<m)
    {
        if(a[i] == b[j] || j==-1)
        {
            i++;
            j++;
        }
        else j=next[j];
    }
    if(j>=m) return i-m+1;
      else return -1;
}
int main()
{

    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
             scanf("%d",&a[i]);
        for(i=0;i<m;i++)
            scanf("%d",&b[i]);
        get_next();
        printf("%d\n",kmp());
    }
    return 0;
}