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

 

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题意：给两个数组a和b，求b在a中出现的的第一个位置，若没有则输出-1，仅仅是数组变成了整型数组，方法与字符数组全然一样。

代码：

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <vector>

#include <set>

#include <queue>

#pragma comment (linker,"/STACK:102400000,102400000")

#define maxn 1000005

#define MAXN 2005

#define mod 1000000009

#define INF 0x3f3f3f3f

#define pi acos(-1.0)

#define eps 1e-6

typedef long long ll;

using namespace std;

int s[maxn],p[maxn],nextval[maxn];

int N,M;

void get_nextval()

{

    int i=0,j=-1;

    nextval[0]=-1;

    while (i<M)

    {

        if (j==-1||p[i]==p[j])

        {

            i++;

            j++;

            if (p[i]!=p[j])

                nextval[i]=j;

            else

                nextval[i]=nextval[j];

        }

        else

            j=nextval[j];

    }

}

int KMP()

{

    int i=0,j=0;

    while (i<N&&j<M)

    {

        if (j==-1||s[i]==p[j])

        {

            i++;

            j++;

        }

        else

            j=nextval[j];

    }

    if (j==M)

        return i-M+1;

    else

        return -1;

}

int main()

{

    int cas;

    scanf("%d",&cas);

    while (cas--)

    {

        scanf("%d%d",&N,&M);

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

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

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

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

        get_nextval();

        int ans=KMP();

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

    }

    return 0;

}

/*

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

*/