Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12326    Accepted Submission(s): 5627

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 <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <iomanip>
using namespace std;
const int INF=0x5fffffff;
const int MS=1000005;
const double EXP=1e-8;
int str1[MS],str2[MS/10];
int next[MS/10];
int N,M;

void get_next(int s[],int next[])
{
    int i,j;
    i=1;j=0;
    next[1]=0;
   // int len=strlen(s);
    int len=M;
    while(i<len)
    {
       if(j==0||s[i-1]==s[j-1])
       {
           i++;
           j++;
           if(s[i-1]==s[j-1])
                next[i]=next[j];
            else
                next[i]=j;
       }
       else
            j=next[j];
    }
}

int KMP(int *s,int *t)
{
    get_next(t,next);
    int i=1;
    int j=1;
    //int len1=strlen(s);
    //int len2=strlen(t);
    int len1=N;
    int len2=M;
    while(i<=len1&&j<=len2)
    {
        if(j==0||s[i-1]==t[j-1])
        {
            i++;
            j++;
        }
        else
            j=next[j];
    }
    if(j>len2)
        return i-len2-1;
    else
        return -1;
}



int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&N,&M);
        for(int i=0;i<N;i++)
            scanf("%d",&str1[i]);
        for(int i=0;i<M;i++)
            scanf("%d",&str2[i]);
        int ans=KMP(str1,str2);
        printf("%d
",ans>=0?ans+1:ans);
    }
    return 0;
}