HDU-Number Sequence -1711

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纯粹模板，没什么好说的

Number Sequence

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 24 Accepted Submission(s) : 14

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

Sample Output

6-1

#include <iostream>
using namespace std;

const int maxt=1000001;
const int maxp=10001;

int p[maxp],t[maxt],nextval[maxp];
int n,m;

void get_nextval()
{
    int i,j;
    nextval[1]=0;
    i=1,j=0;
    while(i<m)
    {
        if(j==0||p[i]==p[j])
        {
            i++;
            j++;
            if(p[i]!=p[j]) nextval[i]=j;
            else nextval[i]=nextval[j];
        }
        else j=nextval[j];
    }
}

int match()
{
    int i,j;
    i=1,j=1;
    while(i<=n&&j<=m)
    {
        if(j==0||t[i]==p[j])
        {
            i++;
            j++;
        }
        else j=nextval[j];
    }
    if(j==m+1) return i-m;
    else return -1;
}

int main()
{
    int ncas,i;
    scanf("%d\n",&ncas);
    while(ncas--)
    {
        scanf("%d %d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%d",&t[i]);
        for(i=1;i<=m;i++)
            scanf("%d",&p[i]);
        get_nextval();
        printf("%d\n",match());
    }
    return 0;
}