Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30288    Accepted Submission(s): 12768


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

這是一道KMP演算法的模板題，KMP演算法在於NEXT陣列的理解，話不多說，記下模板吧。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int a[1000100];
int b[1000100];
int next_[1000010];
void get_next(int m)
{
    int i=0,j=-1;
    next_[0]=-1;
    while(i<m)
    {
        if(j==-1||b[i]==b[j])
        {
            next_[++i]=++j;
        }
        else
        {
            j=next_[j];
        }
    }
}
int kmp(int n,int m)
{
    int i=0,j=0;
    while(i<n)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
            if(j==m)
                return i-m+1;
        }
        else
        {
            j=next_[j];
        }
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        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]);
            }
            get_next(m);
            int ans=kmp(n,m);
            printf("%d\n",ans);
        }
    }

}