hdu 1711Number Sequence

http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5807    Accepted Submission(s): 2608

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<stdio.h>
int F[1000010];
int S[10010];
int n;
int m;
int next[10010];
void get_next()
{
    int i=1;
    int j=0;
    next[1]=0;
    while(i<m)
    {
        if(j==0||S[i]==S[j])
        {
            i++;
            j++;
            next[i]=j;
        }
        else j=next[j];
    }
    
}
int kmp()
{
    int i=1;
    int j=1;
    int flag=1;
    while(i<=n&&j<=m)
    {
        
        if(j==0||F[i]==S[j])
        {
            i++;
            j++;
        }
        else j=next[j];
        if(j>m)
        {
            flag=0;
            printf("%d\n",i-j+1);
            break;
        }
    }
    if(flag) printf("-1\n");
    
}
int main()
{
    int t;
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        for(i=1;i<=n;i++)
        scanf("%d",&F[i]);
        for(i=1;i<=m;i++)
        scanf("%d",&S[i]);
        get_next();
        kmp();
    }
}