HDU 1711 Number Sequence (KMP 入门)

Number Sequence

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

 

题目大意：给定两个序列a，b，求出b序列在a序列中第一次出现的位置，若不在a序列中输出-1；

思路：KMP模板直接套用，这里觉得这位博主写的挺通俗易懂的还是很简单的就看懂了 https://blog.csdn.net/starstar1992/article/details/54913261/

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>

using namespace std;
const int maxn = 1000005;
int rs[4][4], nex[maxn], a[maxn], b[maxn/100 + 5], n, m;
void cal_next(int *str, int len)
{
    nex[0] = -1; int k = -1;
    for (int i = 1; i <= len - 1; i++) {
        while (k > -1 && str[k + 1] != str[i])
            k = nex[k];
        if (str[k + 1] == str[i])k = k + 1;
        nex[i] = k;
    }
}
int KMP(int *a, int la, int  *b, int lb)
{
    cal_next(b , lb);
    int k = -1;
    for (int i = 0; i < la; i++) {
        while (k > -1 && b[k + 1] != a[i])
            k = nex[k];
        if (b[k + 1] == a[i])k = k + 1;
        if (k == lb - 1)return i - lb + 1;
    }
    return -1;
}
int main()
{
    ios::sync_with_stdio(false);
    int T;
    cin >> T;
    while (T--) {
        cin >> n >> m;
        for (int i = 0; i < n; i++)cin >> a[i];
        for (int i = 0; i < m; i++)cin >> b[i];
        int ans = KMP(a, n, b, m);
        if (ans != -1)cout << ans + 1 << endl;
        else cout << "-1" << endl;
    }
    return 0;
}