HDU 1711 Number Sequence (KMP模板)

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 <string.h>
#include <iostream>
#include <cstdio>
#define maxn 1000010
int next[maxn];
int n,m;
void getNext(int* P, int* f)
{
    f[0] = 0;
    f[1] = 0;
    for(int i = 1; i < m; i++)
    {
        int j = f[i];
        while(j && P[i] != P[j])
        {
            j = f[j];
        }
        f[i + 1]=P[i]==P[j]?j+1:0;
    }
}

int kmp(int* T, int* P, int*f)
{
    getNext(P, f);
    int j = 0;
    for(int i = 0; i < n; i++)
    {
        while(j && P[j] != T[i])
        {
            j = f[j];
        }
        if(P[j] == T[i])
        {
            j++;
        }
        if(j == m)
        {
            return i-m+2;
        }
    }
    return -1;
}
int main()
{
    int s1[maxn];
    int s2[maxn];
    int a;
    int t;
    scanf("%d",&t);
    while(t--)
    {scanf("%d%d",&n,&m);
        if(m>n)
        {
            puts("-1");
            continue;
        }
        for(int i=0;i<n;i++)
        {
            scanf("%d",&s1[i]);
        }
        for(int i=0;i<m;i++)
        {
            scanf("%d",&s2[i]);
        }
        int a=kmp(s1,s2,next);
        printf("%d\n",a);

    }
    return 0;
}