HDU1711(KMP)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22739    Accepted Submission(s): 9727

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

 

Source

HDU 2007-Spring Programming Contest

 

//2016.10.7
#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int inf = 0x3f3f3f3f;
int a[1000005], b[10005], nex[10005];

int kmp(int a[], int b[])
{
    int ans = 0;
    nex[0] = -1;
    for(int i = 0, fail = -1; b[i] != inf;)//求nex数组， fail为失配指针
    {
        if(fail==-1 || b[i] == b[fail])
        {
            i++, fail++;
            nex[i] = fail;
        }else fail = nex[fail];
    }
    int i = 0, j = 0;
    for(; a[i] != inf; i++, j++)
    {
        if(j != -1 && b[j] == inf)return i-j+1;
        while(j != -1 && a[i] != b[j])j = nex[j];
    }
    if(b[j] == inf)return i-j+1;
    return -1;
}

int main()
{
    int T, n, m;
    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]);
        a[n] = b[m] = inf;//设置截止符号
        printf("%d
", kmp(a, b));
    }

    return 0;
}