Number Sequence

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

题意描述：

在序列A中找到序列B的位置，若不包含序列B输出-1.

解题思路：

完全kmp算法模板题，先求出序列B的next[]数组，再利用KMP算法求出位置定义一个flag若不包含输出-1.

#include<stdio.h>
#include<string.h>
int a[1000005],b[10005];
int next[10005],n,m;
void get_next(int b[],int m)
{
	int i,j;
	j=0;
	i=1;
	next[0]=0;
	while(i<m)
	{
		
		if(b[i]==b[j])
		{
			next[i]=j+1;
			i++;
			j++;
		}
		if(j==0&&b[i]!=b[j])
		{
			next[i]=0;
			i++;
		}
		if(j>0&&b[i]!=b[j])
			j=next[j-1];
	}
}
int main()
{
	
	int i,j,num,t,flag;
	while(scanf("%d\n",&t)!=EOF)
	{
		while(t--)
		{
			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,m);
			i=0;
			j=0;
			flag=0;
			while(i<=n)
			{
				if(a[i]==b[j])
				{
					j++;
					if(j==m)
					{
						flag=1;
						num=i-m+2;
						break;
					}
					i++;
				}
				if(j==0&&a[i]!=b[j])
					i++;
				if(j>0&&a[i]!=b[j])
					j=next[j-1];
			}
			if(flag==1)
				printf("%d\n",num);
			else
				printf("-1\n");
		}
	}
	return 0;
}