Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11606    Accepted Submission(s): 5294

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

给你s1,s2两个串，让你找到s2在s1中出现的第一个位置。

analyse:

KMP字符串水题，用c++写的，其实真正理解了KMP算法后，你会发现一点都不难。

Time complexity：O(n)　　抓住：主串不回溯

Source code：

// Memory   Time
// 1347K     0MS
// by : Snarl_jsb
// 2014-09-28-14.33
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1000010
#define LL long long
using namespace std;

int n,m;
vector<int> a,b;
vector<int> next;
void GetNext()
{
    int k=0;
    int len=b.size();
    next.push_back(0);
    for(int i=1;i<len;++i)
    {
        while(k!=0&&b[i]!=b[k])
            k=next[k-1];
        if(b[i]==b[k])
            k++;
        next.push_back(k);
    }
//    for(int i=0;i<next.size();++i)
//        cout<<next[i]<<endl;
}
void KMP()
{
    GetNext();
    bool flag=0;
    int asize=a.size(),bsize=b.size(),k=0;
    for(int i=0;i<asize;++i)
    {
        while(k>0&&a[i]!=b[k])
        {
            k=next[k-1];
        }
        if(a[i]==b[k])
            k++;
        if(k==m)
        {
            cout<<i-m+2<<endl;
            flag=1;
            break;
        }
    }
    if(!flag) puts("-1");
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
    int t,tmp;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        a.clear(),b.clear(),next.clear();
        for(int i=0;i<n;++i)
        {
            cin>>tmp;
            a.push_back(tmp);
        }
        for(int i=0;i<m;++i)
        {
            cin>>tmp;
            b.push_back(tmp);
        }
        KMP();
    }
    return 0;
}