hdu 1711---KMP

题目链接

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

 

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
using namespace std;
int n[];
int s[],a[];
int len1,len2;
 
void kmp()
{
    n[]=;
    for(int i=,k=;i<len2;i++)
    {
        while(k>&&a[k]!=a[i]) k=n[k-];
        if(a[k]==a[i]) k++;
        n[i]=k;
    }
}
 
int main()
{
    ///cout << "Hello world!" << endl;
    int T;
    cin>>T;
    while(T--)
    {
        scanf("%d%d",&len1,&len2);
        for(int i=;i<len1;i++)
            scanf("%d",&s[i]);
        for(int i=;i<len2;i++)
            scanf("%d",&a[i]);
        kmp();
        int flag=-;
        for(int i=,j=;i<len1;i++)
        {
            while(j>&&a[j]!=s[i]) j=n[j-];
            if(a[j]==s[i]) j++;
            if(j==len2) { flag=i-len2+; break; }
        }
        if(flag>=) printf("%d\n",flag+);
        else puts("-1");
    }
    return ;
}