hdu 1711

                                Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8122 Accepted Submission(s): 3686

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
  

Source

HDU 2007-Spring Programming Contest

Recommend

lcy

代码：

#include<iostream>
using namespace std;

const int MAX=1000000+10;
int a[MAX];
int b[10000+10];
int next[10000+10];

void get_next(int *t,int len)
{
    int i=0,j=-1;
    next[0]=-1;
    int Lt=len;
    while(i<Lt)
    {
        if(j==-1||t[i]==t[j])
        {
            ++i,++j;
            next[i]=j;            
        }
        else
        j=next[j];
    }
}

int KMP(int a[],int b[],int lena,int lenb)
{
    int i=0,j=0;
    while (i<lenb && j<lena)
    {
        if (j == -1 || a[j] == b[i])++i,++j;
        else j=next[j];
    }
    if (j == lena)return i-j+1;
    return -1;
}

int main()
{
    int n,m,t;
    cin>>t;
    while (t--)
    {
        cin>>n>>m;
        for (int i=0;i<n;++i)
		scanf("%d",&a[i]);
        for (int i=0;i<m;++i)
		scanf("%d",&b[i]);
        get_next(b,m);
        cout<<KMP(b,a,m,n)<<endl;
    }
    return 0;
}