hdu 1711

                                    Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24293    Accepted Submission(s): 10314

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 板子题 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<map>
#include<set>
#include<vector>
#include<cstdlib>
#include<string>
#define eps 0.000000001
typedef long long ll;
typedef unsigned long long LL;
using namespace std;
const int N=1000000+100;
int a[N],b[N];
int m,n;
void get_next(int T[],int next[]){
    int j=0,k=-1;
    next[0]=-1;
    while(j<n){
        if(k==-1||T[j]==T[k]){
            j++;
            k++;
            next[j]=k;
            //cout<<k<<" ";
        }
        else
            k=next[k];
    }
}
void  kmp(int S[],int T[]){
    int i=0,j=0;
    int next[100000];
    get_next(T,next);
    while(i<m){
        if(j==-1||S[i]==T[j]){
            i++;
            j++;
        }
        else
            j=next[j];
        if(j==n){
            printf("%d
",i-n+1);
            return;
        }
    }
    printf("-1
");
}
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&m,&n);
        int x,y;
        for(int i=0;i<m;i++){
            scanf("%d",&a[i]);
        }
        for(int i=0;i<n;i++){
            scanf("%d",&b[i]);
        }
        //get_next(b);
        kmp(a,b);
    }
}