HDU 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 33233    Accepted Submission(s): 13890

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

#include<bits/stdc++.h>
using namespace std;
int text[1000003],p[10003],n,m,next1[10003];
void NEXT()
{
    int k=-1,q=0;
    memset(next1,-1,sizeof(next1));
    while(q<m)
    {
        if(k==-1||p[k]==p[q])
            k++,q++,next1[q]=k;
        else k=next1[k];
    }
}
int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        int i,j;
        for(i=0;i<n;i++)
            scanf("%d",&text[i]);
        for(i=0;i<m;i++)
            scanf("%d",&p[i]);
        NEXT();
        int s=0,q=0,t=-1;
        while(s<n)
        {
            if(q==-1||p[q]==text[s])
                q++,s++;
            else q=next1[q];
            if(q==m)
            {
                t=s-m+1;break;
            }
        }
        printf("%d\n",t);
    }
    return 0;
}