Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 42790    Accepted Submission(s): 17672

 

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

 

终于理解KMP了，开心

顺手敲了模板题

https://blog.csdn.net/v_JULY_v/article/details/7041827#t17

KMP详解超好

#include<bits/stdc++.h>
using namespace std;
int txt[1000005];
int net[10005];
int mode[10005];
void GetNext(int p[],int net[],int pLen)
{
    //int pLen = strlen(p);
    net[0]=-1;
    net[1]=0;
    int len=0;
    for(int i=2; i<pLen; i++)
    {
        if(p[i-1]==p[len])
        {
            net[i]=net[i-1]+1;
            len++;
        }
        else
        {
            net[i]=0;
            len=0;
        }
    }
}
int KMP(int text[],int mode[],int net[],int len1,int len2)
{
    //  int len1=strlen(text);
    // int len2=strlen(mode);
    int i=0,j=0;
    int l=0;
    while(i<len1&&j<len2)
    {



        if(text[i]==mode[j]||j==-1)
        {
            i++;
            j++;

        }
        else
        {

            j=net[j];


        }

    }
    if(j==len2)
    {
        return i-len2+1;
    }
    else
        return -1;
}

int main()
{

    int t;
    cin>>t;

    while(t--)
    {
        int lena,lenb;
        scanf("%d%d",&lena,&lenb);
        for(int i=0; i<lena; i++)
        {
            scanf("%d",&txt[i]);

        }
        for(int i=0; i<lenb; i++)
        {
            scanf("%d",&mode[i]);
        }

        GetNext(mode,net,lenb);
        printf("%d\n",KMP(txt,mode,net,lena,lenb));
    }
    // next[0]=0;
    // int net[1005];

    return 0;
}

/*

13
13 5
1 1 3 4 5 6 1 1 3 4 5 6 5
3 4 5 6 5

*/