Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15017    Accepted Submission(s):
6585

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

#include<stdio.h>
#include<string.h>
#define MAX 10010
int n,m;
int b[MAX],a[MAX*100];
int next[MAX];
void getmap()
{
	int i;
	for(i=0;i<n;i++)
	scanf("%d",&a[i]);
	for(i=0;i<m;i++)
	scanf("%d",&b[i]);
}
void getfail()
{
	int i,j;
	next[0]=next[1]=0;
    for(i=1;i<m;i++)
    {
    	j=next[i];
    	while(j&&b[i]!=b[j])
    	    j=next[j];
    	next[i+1]=b[i]==b[j]?j+1:0;
	}
}
void kmp()
{
	int i,j=0;
	int ok=0;
	for(i=0;i<n;i++)
	{
		while(j&&a[i]!=b[j])
		    j=next[j];
		if(a[i]==b[j])
		    j++;
		if(j==m)
		{
			ok=1;
			printf("%d\n",i-m+2);//难点在这里
			return ;
		}
	}
	if(!ok)
	printf("-1\n");
}
int main()
{
	int i,t;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		getmap();
		getfail();
		kmp();
	}
	return 0;
}