Number Sequence（kmp算法）

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. 

InputThe 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]. 
OutputFor 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<cstdio>
#include<cstring>
#include<string>
#include<set>
#include<vector>
#include<stack>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<algorithm>
#include<functional>
#include<sstream>
using namespace std;
const double g = 10.0, eps = 1e-9;
const int N = 1000000 + 5, maxn = 10000 + 5, inf = 0x3f3f3f3f;

int ext[N], str[maxn], ptr[N];
void getnext(int slen)
{
    ext[0] = -1;
    int k = -1;
    for (int i = 1; i <= slen - 1; i++)
    {
        while (k>-1 && str[k + 1] != str[i])k = ext[k];
        if (str[k + 1] == str[i])k++;
        ext[i] = k;
    }
}
int kmp(int plen, int slen)
{
    int k = -1;
    for (int i = 0; i <= plen - 1; i++)
    {
        while (k>-1 && str[k + 1] != ptr[i])k = ext[k];
        if (str[k + 1] == ptr[i])k++;
        if (k == slen - 1)return i - slen + 2;
    }
    return -1;
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    //   cout<<setiosflags(ios::fixed)<<setprecision(2);
    int t, n, m;
    cin >> t;
    while (t--) {
        cin >> n >> m;
        for (int i = 0; i<n; i++)cin >> ptr[i];
        for (int i = 0; i<m; i++)cin >> str[i];
        getnext(m);
        cout << kmp(n, m) << endl;
    }
    return 0;
}