Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6111    Accepted Submission(s): 2744

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

//kmp，裸题
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <cstdlib>
using namespace std;
int b[10005],a[1000010];
int next[10005];
int m,n;
void get()
{
    memset(next,0,sizeof(next));
    int i,j;
    i = 0;
    j = -1;
    next[0] = -1;
    while(i<m)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;
            j++;
            next[i] = j;
        }
        else
            j = next[j];
    }
}
int kmp()
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
        {
            j = next[j];
        }
    }
    if(j==m)
        return i-j+1;
    else
        return -1;
}
int main()
{
    int i,j,k,T;
    scanf("%d",&T);
    while(T--)
    {
        memset(b,0,sizeof(b));
        memset(a,0,sizeof(a));
        scanf("%d%d",&n,&m);
        for(i=0;i<n;i++)
            scanf("%d",a+i);
        for(i=0;i<m;i++)
            scanf("%d",b+i);
        get();
        //cout<<next[0]<<endl;
        //system("pause");
        int pos = kmp();
        printf("%d\n",pos);
    }
    return 0;
}