Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30523    Accepted Submission(s): 12849

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

基础的kmp字符串匹配

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;

int k[10010];
int h[1000010];
int NEXT[10010];

void GetNext(int k[],int y,int NEXT[]){
    int klen=y+1;
    int i=0;
    int j=-1;
    NEXT[0]=-1;
    while(i<klen){
        if(j==-1||k[i]==k[j]){
            ++i;
            ++j;
            NEXT[i]=j;
        }else{
            j=NEXT[j];
        }
    }
}

int kmp_find(int h[],int k[],int x,int y,int NEXT[]){
     GetNext(k,y,NEXT);
    int i=0;
    int j=0;
    int hlen=x+1;
    int klen=y+1;
    while(i<hlen&&j<klen){
        if(j==-1||h[i]==k[j]){
            i++;j++;
        }else{
            j=NEXT[j];
        }
    }
     if(j==klen){
            return i-j+1;
        }
    return -1;
}
int main(){
    int a;
    scanf("%d",&a);
    while(a--){
        int n,m;
        scanf("%d%d",&n,&m);
        int x,y;
        for(int i=0;i<n;i++){
          scanf("%d",&h[i]);
            x=i;
        }
        for(int i=0;i<m;i++){
          scanf("%d",&k[i]);
            y=i;
        }

        int t=kmp_find(h,k,x,y,NEXT);
        cout<<t<<endl;
        memset(NEXT,0,sizeof(NEXT));
        memset(h,0,sizeof(h));
        memset(k,0,sizeof(k));
    }
return 0;
}