【KMP】Number Sequence（含KMP解说视频链接）_sequence-number indicator-优快云博客

本文深入解析KMP算法在字符串匹配中的应用，通过实例演示如何寻找子串在主串中的首次出现位置，包括KMP算法的预处理过程和匹配流程，适合初学者理解和掌握。

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问题：Number Sequence

题目描述

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.

输入

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].

输出

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

样例输入

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

样例输出

6
-1

题目意思就是给定AB两个长度为n,m的字符串（A长B短）进行匹配，在A中找到B第一次出现的下标（完全匹配），若不存在则输出-1.

刚开始搜网络上的方法，大家都使用KMP的算法，一开始没有看懂KMP的预处理是怎么进行的。

后来看到一个印度小哥的解说视频就很好理解了

https://www.bilibili.com/video/av47471886/

#include <iostream>
using namespace std;
int nex[10001];
int a[1000001], b[10001];
int n, m;
int kmp(int *a,int n,int *b,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 = nex[j];
	}
	return -1;
}

void getnext(int *b,int m) {
	int i = 0,j = -1;
	nex[0] = -1;
	while (i < m) {
		if (j == -1 || b[i] == b[j])nex[++i] = ++j;
		else j = nex[j];
	}
}
int main()
{
	int t;
	cin >> t;
	while (t--) {
		int n, m;
		cin >> n >> m;
		for (int i = 0; i < n; i++)cin >> a[i];
		for (int i = 0; i < m; i++)cin >> b[i];
		getnext(b,m);
		cout << kmp(a,n,b,m)<<endl;
	}
	return 0;
}