Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 26895 Accepted Submission(s): 11360

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

 
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

-1

http://blog.csdn.net/v_july_v/article/details/7041827 這篇博客kmp寫的很詳細（大牛的博客）。

KMP的模板題：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define maxx 10100
using namespace std;
int a[maxx*100];
int b[maxx];
int nex[maxx];
int n,m;
void prekmp()
{
    int p=0,q=-1;nex[0]=-1;
    while(p<m)
    {
        while(q!=-1&&b[p]!=b[q])
            q=nex[q];
        nex[++p]=++q;
    }
}
int kmp()
{
    prekmp();
    int p=0,q=0;
    while(p<n&&q<m)
    {
        while(q!=-1&&a[p]!=b[q])
            q=nex[q];
        p++;q++;
        if(q==m)return p-m+1;
    }
    return -1;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
            scanf("%d",a+i);
        for(int i=0;i<m;i++)
            scanf("%d",b+i);
        printf("%d\n",kmp());
    }
}

注意！

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