hdu1711 Number Sequence 求模式串在主串中的位置

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<cstring>
#include<cstdio>
#include<string>
using namespace std;
int s[1000010],t[10010];//s为主串，t为模式串
int nextx[10010];//从1开始
int slen,tlen;
void get_next()////得到模式串str1的next[];
{
    int i=0,j=-1;
    nextx[0]=-1;
    while(i<tlen)
    {
        if(j==-1||t[i]==t[j])
            nextx[++i]=++j;
        else
            j=nextx[j];
    }
}
int kmp_index()////返回首次出现的位置,从0开始
{
    int i=0,j=0;
    while(i<slen&&j<tlen)
    {
        if(j==-1||s[i]==t[j])
        {
            i++;
            j++;
        }
        else
            j=nextx[j];
    }
    if(j==tlen)
        return i-tlen+1;
    else
        return -1;//匹配不成功
}
int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        cin>>slen>>tlen;
        for(int i=0;i<slen;i++)
            cin>>s[i];
        for(int i=0;i<tlen;i++)
            cin>>t[i];
        get_next();
        cout<<kmp_index()<<endl;
    }
    return 0;
}