Number Sequence （HDU 1711）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24116    Accepted Submission(s):
10232

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

 

#include<iostream>
#include<cstring>
using namespace std;
int a[],b[],l1,l2;
int Case,ne[];
void get_ne()
{
    int j=,k=-;
    ne[]=-;
    while(j<l2)
    {
        if(k==-||b[j]==b[k])
            ne[++j]=++k;
        else k=ne[k];
    }
}
void kmp()
{
    int i=,j=;
    get_ne();
    while(i<l1)
    {
        if(j==-||a[i]==b[j])i++,j++;
        else j=ne[j];
        if(j==l2) {cout<<i-l2+<<endl;return;}
    }
    cout<<-<<endl;
    return;
}
int main()
{
    cin>>Case;
    while(Case--)
    {
        memset(a,,sizeof(a));
        memset(b,,sizeof(b));
        cin>>l1>>l2;
        for(int i=;i<l1;i++)cin>>a[i];
        for(int j=;j<l2;j++)cin>>b[j];
        kmp();
    }
}