HDU1711(KMP入门题)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 17971    Accepted Submission(s): 7854

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<cstdio>
#include<cstring>
using namespace std;
const int MAXN=;
int a[MAXN],b[MAXN],next[MAXN];
int lena,lenb;
void getnext()
{
    int i=,j=-;
    next[]=-;
    while(i<lenb)
    {
        if(j==-||b[i]==b[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else    j=next[j];
    }
}
int KMP()
{
    getnext();
    int i=,j=;
    while(i<lena&&j<lenb)
    {
        if(j==-||a[i]==b[j])
        {
            i++;
            j++;
        }
        else    j=next[j];
    }
    if(j==lenb)    return i-j+;
    else    return -;
}
int main()
{
   int T;
   scanf("%d",&T);
   while(T--)
   {
           memset(next,,sizeof(next));
           scanf("%d%d",&lena,&lenb);
           for(int i=;i<lena;i++)
               scanf("%d",&a[i]);
           for(int i=;i<lenb;i++)
               scanf("%d",&b[i]);
           printf("%d\n",KMP());
   }

    return ;
}