AC日记——Number Sequence hdu 1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24928    Accepted Submission(s):
10551

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

 

 

思路：

　　kmp模板；

 

 

来，上代码：

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

const int N = ;
int nest[],n,m;
int s[],t[];
int slen,tlen;

void getnest()
{
    memset(nest,,sizeof(nest));
    int j,k;
    j=;k=-;nest[]=-;
    while(j<tlen)
    {
        if(k==-||t[j]==t[k])
            nest[++j]=++k;
        else k=nest[k];
    }
}

int kmp_index()
{
    int i=;int j=;
    getnest();
    while(i<slen&&j<tlen)
    {
        if(j==-||s[i]==t[j])
        {
            i++;j++;
        }
        else j=nest[j];
    }
    if(j==tlen) return i-tlen+;
    else return -;
}

int kmp_count()
{
    int ans=;
    int i,j=;
    if(slen==&&tlen==)
    {
        if(s[]==t[]) return ;
        else return ;
    }
    getnest();
    for(i=;i<slen;i++)
    {
        while(j>&&s[i]!=t[j])
        {
            j=nest[j];
        }
        if(s[i]==t[j]) j++;
        if(j==tlen)
        {
            ans++;
            j=nest[j];
        }
    }
    return ans;
}

int main()
{
    int TT;
    int i,cc;
    cin>>TT;
    while(TT--)
    {
        cin>>n>>m;
        for(int i=;i<n;i++)
            cin>>s[i];
        for(int j=;j<m;j++)
            cin>>t[j];
        tlen=m,slen=n;
        cout<<kmp_index()<<endl;
    }
    return ;
}