Number Sequence

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

Total Submission(s): 40388    Accepted Submission(s): 16659

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

给出两个数组，问第二个数组是不是第一个数组的子序列，如果是，输出第二个数组的第一个元素在第一个数组中的位置，否则输出-1

AC代码

#include <bits/stdc++.h>

#define ll long long

#define ms(a) memset(a,0,sizeof(a))

using namespace std;

const int maxn=1e6+10;

int a[maxn],b[maxn];

int Next[maxn];

int  n,m;

// Next数组下标从1开始

void getNext()

{

	int j,k;

	j=0;k=-1;Next[0]=-1;

	while(j<m)

	{

		if(k==-1||b[j]==b[k])

			Next[++j]=++k;

		else

			k=Next[k];

	}

}

/*

返回第二个数组在第一个数组中首次出现的位置

返回的位置是从0开始的。

*/

int KMP_Index()

{

	int i=0,j=0;

	getNext();

	while(i<n&&j<m)

	{

		if(j==-1||a[i]==b[j])

		{

			i++;j++;

		}

		else

			j=Next[j];

	}

	if(j==m)

		return i-m;

	else

		return -1;

}

int main(int argc, char const *argv[])

{

	int t;

	scanf("%d",&t);

	while(t--)

	{

		scanf("%d%d",&n,&m);

		ms(a);

		ms(b);

		ms(Next);

		for(int i=0;i<n;i++)

			scanf("%d",&a[i]);

		for(int i=0;i<m;i++)

			scanf("%d",&b[i]);

		if(KMP_Index()!=-1)

			printf("%d\n",KMP_Index()+1);

		else

			printf("%d\n",KMP_Index());

	}

	return 0;

}