裸kmp算法

Number Sequence

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

 

刚接触kmp还不是很理解，next失效函数

 

#include <iostream>
#include <cstdio>
using namespace std;
int f[10001];
int n,m;
int a[1000001],b[10001];
int kmp(int t[],int p[])
{
    int i=0,j=0;
    while(i<n&&j<m)
    {
        if(j==-1||t[i]==p[j])
        {
            i++;
            j++;
        }
        else
            j=f[j];
    }
    if(j>=m)
        return i-m+1;
    return -1;
}
void next(int p[])
{
    int k=-1,j=0;
    f[0]=-1;
    while(j<m)
    {
        if(k==-1||p[k]==p[j])
        {
            k++;
            f[++j]=k;
        }
        else
            k=f[k];
    }
}
int main()
{
    int T,i,j;
    //freopen("in.txt","r",stdin);
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(i=0;i<n;i++)
            cin>>a[i];
        for(i=0;i<m;i++)
            cin>>b[i];
        next(b);
        cout<<kmp(a,b)<<endl;
    }
}