HDU 1711

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;

}