F - Number Sequence

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

题意概括  :

给出两个数组，判断第二个数组是否包含在第一个数组里，如果是，则输出第二个数组中第一个字符在第一个字符串中出现的位置。

解题思路  :

首先求出第二个数组的next数组，然后与第一个数组进行匹配，当第二个数组的下标等于第二个数组长度时结束循环，或者当第一个数组找到最后一个时结束循环，如果找到了第二个数组，则用当前第一个数组的下标的值减去第二个数组的长度然后加一。找不到输出-1。

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

int a[1000010],b[10010],next[10010];
int n,m;

void get_next(int b[])
{
	int i,j;
	i = 1;
	j = 0;
	next[0] = 0;
	
	while(i < m)
	{
		if(j == 0&&b[i] != b[j])
		{
			i ++;
		}
		if(j > 0&&b[i] != b[j])
		{
			j = next[j-1];
		}
		if(b[i] == b[j])
		{
			next[i] = j+1;
			i ++;
			j ++;
		}
	}
}
int KMP(int a[],int b[])
{
	int i,j,find;
	i = 0;
	j = 0;
	find = -1;
	
	while(i <= m&&j < n)
	{
		if(i == 0&&b[i] != a[j])
		{
			j ++;
		}
		if(i > 0&&b[i] != a[j])
		{
			i = next[i-1];
		}
		if(b[i] == a[j])
		{
			i ++;
			j ++;
		}
		if(i == m)
		{
			find = j-m+1;
			break;
		}
	}
	return find;
}
int main()
{
	int T,k,i,j;
	
	scanf("%d",&T);
	while(T --)
	{
		memset(next,0,sizeof(next));
		scanf("%d %d",&n,&m);
		for(i = 0;i < n;i ++)
		{
			scanf("%d",&a[i]);
		}
		for(j = 0;j < m;j ++)
		{
			scanf("%d",&b[j]);
		}
		get_next(b);
		k = KMP(a,b);
		printf("%d\n",k);
	}
	return 0;
}