HDU 1711 Number Sequence（KMP）

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

 

思路：很明显的kmp题

#include<bits/stdc++.h>
using namespace std;
#define ll long long

const ll inf = 2000000000000;
const int mod = 1000000007;
const int maxn = 1000 + 8;

int n, m, k, t, ne[10000 + 8];

int a[1000000 + 8], b[10000 + 8];

void getnext()
{
    ne[0] = ne[1] = 0;
    for(int i = 1; i < m; i++)
    {
        int j = ne[i];///
        while(j && b[i] != b[j])
            j = ne[j];
        ne[i + 1] = b[i] == b[j] ? j + 1 : 0;///如果两个字母相同，则i和j都加加；否则j = 0，重新比较
    }
}

int kmp()
{
    int j = 0;
    for(int i = 0; i < n; i++)
    {
        while(j && b[j] != a[i])///如果匹配不成功
            j = ne[j];
        if(a[i] == b[j])
           j++;
        if(j == m)
            return i - m + 2;
    }
    return -1;
}

int main()
{
    scanf("%d", &t);
    while(t--)
    {
        memset(ne, 0, sizeof(ne));
        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]);
        getnext();
        k = kmp();
        printf("%d\n", k);
    }
    return 0;
}