Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

/*KMP算法关键是求next函数，第一道KMP题，最基础的题，
借鉴普浩冉的代码和讲解，还算有点理解吧= =
*/
#include<stdio.h>
int a[1000002];
int s[10002],next[10002];
int m,n;
void get_next(int *s)
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<m-1){
		if(j==-1||s[i]==s[j])
		{
			i++;
			j++;
			next[i]=j;
		}
		else
		j=next[j];
	}
}
int kmp(int *a,int *s)
{
	int i=0,j=0;
	get_next(s);
	while(i<=n-1)
	{
		if(j==-1||a[i]==s[j])
		{
			i++;
			j++;
		}
		else
		    j=next[j];
		if(j==m)
		    return i-m+1;
	}
	return -1;
}
int main()
{
	int test,i;
	scanf("%d",&test);
	while(test--)
	{
		scanf("%d %d",&n,&m);
		for(i=0;i<n;i++)
		    scanf("%d",&a[i]);
		for(i=0;i<m;i++)
		    scanf("%d",&s[i]);
		printf("%d\n",kmp(a,s));
	}
	return 0;
}