hdu1711 （Number Sequence） KMP模板

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模板：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define N 1000006
#define M 10004
int a[N],b[M];
int n,m;
int next[M];
int t;
void Get_next()
{
    int 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()
{
    Get_next();
    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-m+1;
    else
        return -1;
}

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        memset(next,-1,sizeof(next));
        scanf("%d%d",&n,&m);
        int i,j;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(j=0;j<m;j++)
            scanf("%d",&b[j]);
        printf("%d\n",KMP());
    }
    return 0;
}