Number Sequence +KMP

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 14

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
 
 
/*
题目大意：给你两组数据，判断第二组是否存在于第一组，如果存在从第几位数开始，
不存在则输出-1。
思路：直接用KMP算法，找到第一个匹配的地方，然后将i-m+1及时第一次匹配的位置
*/
#include<iostream>
#include<cstdio>
using namespace std ;
#define MAX 1000005 
int next[MAX],m,n;
int a[MAX] , b[MAX];
void pre_next(int k )
{
	int i = 0 , j = -1 ;
	next[0] = - 1 ;
	while( i < k )
	{
		if( j == -1 || a[i] == a[j])
		{
			i++ ;
			j++ ;
			next[i] = j ;
		}
		else
			j = next[j] ;
	}
}

int  KMP()
{
	int i = 0 , j = 0 ,count = 0 ;
	while( i <= n )
	{
		if(j== - 1 || a[j]==b[i]){
			i++;
			j++ ;
		}
		else
			j = next[j] ;
		if(j == m ){
			return  i-m+1;
		}
	}
	return -1 ;
}

int main(void)
{	
	int t ;

	scanf("%d",&t);
	while(t--)
	{
		int i ;
		scanf("%d %d",&n,&m);
		for( i = 0 ; i < n ; i ++)
			scanf("%d",&b[i]);
		for( i = 0 ; i < m ; i ++)
			scanf("%d",&a[i]);
		pre_next(m);
		if(n<m)
			printf("-1\n");
		else
			printf("%d\n",KMP());

	}
	return 0 ; 
}