題目

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 40385    Accepted Submission(s): 16656



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
   

程式碼

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN=1e6 + 10;
int a[MAXN],b[MAXN];
int nxt[MAXN];
int n,m;
void getnxt()
{
    int
 
 i=1;
    int j=0;
    nxt[0]=0;
    while(i<m)
    {
        if(b[i]==b[j])
        {
            nxt[i++]=++j;
        }
        else if(!j)
        {
            i++;
        }
        else
        {
            j=nxt[j-1];
        }
    }
}
int kmp()
{
    int i=0;
    int j=0;
    while(i<n && j<m)
    {
        if
 
(a[i]==b[j])
        {
            i++;
            j++;
        }
        else if(!j)
        {
            i++;
        }
        else
        {
            j=nxt[j-1];
        }
    }
    if(j==m) return i-m+1;
    else return -1;
}
int main()
{
    int t,i,k;  
    scanf("%d",&t);
    while(t--)
    {
        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]);
        getnxt();
        cout<<kmp()<<endl;
    }
    return 0;
}