Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22568    Accepted Submission(s): 9639

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算法还需多理解。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;

int n,m;
int N[1000005],M[10005],Pi[10005];

void preFix()
{
    memset(Pi,0,sizeof(Pi));
    int k=0;
    for(int q=2;q<=m;q++)
    {
        while(k>0&&M[k+1]!=M[q])
            k=Pi[k];
        if(M[k+1]==M[q])
            k++;
        Pi[q]=k;
    }
}

int KMP()
{
    preFix();
    int q=0;
    for(int i=1;i<=n;i++)
    {
        while(q>0&&M[q+1]!=N[i])
            q=Pi[q];
        if(M[q+1]==N[i])
            q++;
        if(q==m)
            return i-m+1;
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&N[i]);
        for(int j=1;j<=m;j++)
            scanf("%d",&M[j]);
        int ans=KMP();
        printf("%d\n",ans);

    }
    return 0;
}