Number Sequence HDU - 1711 （KMP模板题）

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
代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int maxn = 1e6 + 10;
 4 int n, m;
 5 int a[maxn];
 6 int b[maxn];
 7 int nex[maxn];
 8 
 9 void getnext()
10 {
11     int i = 1, j = 0;
12     nex[1] = 0;
13     while (i < m)
14     {
15         if (j == 0 || b[i] == b[j])
16         {
17             i++, j++;
18             if (b[i] != b[j]) nex[i] = j;
19             else nex[i] = nex[j];
20         }
21         else j = nex[j];
22     }
23 }
24 
25 int kmp()
26 {
27     int i, j; i = j = 1;
28     while (i <= n && j <= m)
29     {
30         if (j == 0 || a[i] == b[j]) i++, j++;
31         else j = nex[j];
32     }
33     if (j > m)
34         return i - m;
35     else return -1;
36 }
37 
38 int main()
39 {
40     int T; cin >> T;
41     while (T--)
42     {
43         cin >> n >> m;
44         for (int i = 1; i <= n; i++)
45             scanf("%d", a + i);
46         for (int i = 1; i <= m; i++)
47             scanf("%d", b + i);
48         getnext();
49         int res = kmp();
50         cout << res << endl;
51     }
52 }