hdu 1711 Number Sequence-优快云博客

本文链接：https://blog.youkuaiyun.com/zhang__liuchen/article/details/78502308

本文介绍了一种使用KMP算法解决特定数值序列匹配问题的方法。给定两个数列，任务是找到第一个数列中是否存在与第二个数列完全相同的子序列及起始位置。通过构造跳转表(next数组)来加速匹配过程，并详细展示了如何实现这一算法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31372 Accepted Submission(s): 13170

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

KMP算法，得到跳转表（next数组），然后进行匹配。

#include <iostream>
#include <cstring>
#include <cstdio>
const int M=1e4;
const int N=1e6;
int Next[M+10];
int str[N+10];
int mo[M+10];
int n,m;
using namespace std;
void getNext()
{
	Next[0]=-1;
	int i=0,j=-1;
	while(i<m)
	{
		if(j==-1||mo[i]==mo[j])
		{
			i++;
			j++;
			Next[i]=j;
		}
		else
			j=Next[j];
	}
}
int main()
{
	int T;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++)
			scanf("%d",&str[i]);
		for(int i=0;i<m;i++)
			scanf("%d",&mo[i]);
		getNext();
		int ans=-1;
		int i=0,j=0;
		while(i<n)//KMP匹配
		{
			if(j==-1||str[i]==mo[j])
			{
				i++;
				j++;
			}
			else
				j=Next[j];
			if(j==m)
			{
				ans=i-j+1;
				break;
			}
		}


		printf("%d\n",ans);
	}
	return 0;
}