HDOJ1711(Number Sequence)

Number Sequence

Time Limit: 1000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 768    Accepted Submission(s): 278

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

 

 

/**//*
    2009-05-10 11:04:54 Accepted 1711 515MS 4192K 944 B C++ 
    试着把自己做的模板用一下 
*/ 
#include <iostream>
#include <cstdio>
using namespace std;

const int M = 10004;
const int N = 1000004;

int S[N],//主串    
    T[M];//模式串
int Slen, Tlen;//主串和模式串的长度
int next[M];

void get_next()
{
    int j, k;

    j = 0; k = -1; next[0] = -1;

    while(j < Tlen)
        if(k == -1 || T[j] == T[k])
        {
            next[++j] = ++k;
        }
        else
            k = next[k];

}

int KMP_Index()
{//此函数用于返回模式串在主串中第一次出现的位置
 //不存在返回-1
    int i, j;
    i = j = 0;

    get_next();

    while(i < Slen && j < Tlen)
        if(j == -1 || S[i] == T[j])
        {
            i++; j++;
        }
        else
            j = next[j];

    if(j == Tlen)
        return i - Tlen;
    else
        return -2;

}

int main()
{
    int n, i;
    scanf("%d", &n);
    while(n--)
    {
        scanf("%d%d", &Slen, &Tlen);
        for(i = 0; i < Slen; i++)
            scanf("%d", S + i);
        for(i = 0; i < Tlen; i++)
            scanf("%d", T + i);
            
        printf("%d\n", KMP_Index() + 1);
    }
    return 0;
}