Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25500    Accepted Submission(s): 10771

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

 

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#include<iostream>
#include<math.h>
#include<memory.h>
#include<stdlib.h>
using namespace std;
int a[1000005],b[10001],n,m;
void GetNext(int b[],int next[])
{
	int j=0,k=-1;
	next[0]=-1;
	while(j<m)
	{
		if(k==-1||b[j]==b[k])
		{
			j++; k++;
			if(b[j]!=b[k])
				next[j]=k;
			else
				next[j]=next[k];
		}
		else
			k=next[k];
	}
}
int KMP(int a[],int b[])
{
	int next[m],i=0,j=0;
	GetNext(b,next);
	while(i<n&&j<m)
	{
		if(j==-1||a[i]==b[j])
		{
			i++; j++;
		}
		else
			j=next[j];
	}
	if(j>=m)
		return (i-m+1);
	else
		return -1;
}
int main()
{
	int num;
	cin>>num;
	while(num--)
	{
		int i,temp;
		cin>>n>>m;
		for(i=0;i<n;i++)
			cin>>a[i];
		for(i=0;i<m;i++)
			cin>>b[i];
		temp=KMP(a,b);
		cout<<temp<<endl;	
	}
}