HDU 1711 Number Sequence

题目链接：HDU1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19372    Accepted Submission(s): 8328

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
   
   
   
   

 

题意：两个串s和t每个串元素由数字组成，问第一个串是不是包含第二个串。

题目分析：KMP模板

//
//  main.cpp
//  HDU1711
//
//  Created by teddywang on 16/4/21.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int s[1000010],t[10010],nexts[100020];
int T,m,n;

void getnext()
{
    int i=0,j=-1;
    nexts[0]=-1;
    while(i<m)
    {
        if(j==-1||t[i]==t[j])
        {
            //nexts[++i]=++j;
            if(t[++i]!=t[++j])
                nexts[i]=j;
            else nexts[i]=nexts[j];
        }
        else j=nexts[j];
    }
}

int kmp()
{
    int i=0,j=0;
    int ans=-1;
    while(i<n&&j<m)
    {
        if(s[i]==t[j]||j==-1)
        {
            if(j==m-1)
                ans=i-m+2;
            ++i;++j;
        }
        else j=nexts[j];
    }
    return ans;
}

int main()
{
    cin>>T;
    while(T--)
    {
        cin>>n>>m;
        for(int i=0;i<n;i++)
            cin>>s[i];
        for(int i=0;i<m;i++)
            cin>>t[i];
        getnext();
        cout<<kmp()<<endl;
    }
}