HDOJ 1711:Number Sequence KMP匹配

最新推荐文章于 2020-02-25 18:36:37 发布

bruce128

最新推荐文章于 2020-02-25 18:36:37 发布

阅读量1.5k

点赞数 1

CC 4.0 BY-SA版权

分类专栏： OnlineJudge 文章标签：算法 url im

本文链接：https://blog.youkuaiyun.com/bruce128/article/details/7355790

OnlineJudge 专栏收录该内容

22 篇文章

订阅专栏

本文介绍如何使用KMP算法解决序列匹配问题，提供了一段AC代码，并详细解释了算法实现过程。针对给定的两个序列，找到使得后一序列成为前一序列子序列的最小索引。

题目URL：http://acm.hdu.edu.cn/showproblem.php?pid=1711

这道题目是单纯的KMP算法，而且不用改进的KMP就可以通过。没有什么特殊要注意的细节。

我的AC代码：

#include <stdio.h>

const int Max = 1000000 + 10;
int a[Max], b[Max], next[Max];
int cases, m, n;


void getNext()
{
	next[0] = -1, next[1] = 0;
	int i = 1, j = 0;
	while(i < n)
	{
		if(j == -1 || b[j] == b[i])
		{
			++i, ++j;
			next[i] = j;
		}
		else j = next[j];
	}
}

int kmp()
{
	int i(0), j(0);

	while(j != n && i < m)
	{
		if(j == -1 || a[i] == b[j]) ++i, ++j;
		else j = next[j];
	}
	if(j == n) return i - n + 1;
	else return -1;
}

int main()
{	
	int result;
	scanf("%d", &cases);

	while(cases --)
	{
		scanf("%d%d", &m, &n);
		for(int i(0); i<m; i++) scanf("%d", a+i);
		for(int i(0); i<n; i++) scanf("%d", b+i);

		getNext();
		result = kmp();
		printf("%d\n", result);
	}
	return 0;
}

原题如下：

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16886 Accepted Submission(s): 7420

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

Recommend

lcy