HDU 1711 Number Sequence

最新推荐文章于 2020-07-12 20:18:12 发布

原创最新推荐文章于 2020-07-12 20:18:12 发布 · 295 阅读

0 ·

CC 4.0 BY-SA版权

KMP 专栏收录该内容

10 篇文章

订阅专栏

本文介绍了一种使用KMP算法解决特定序列匹配问题的方法。该问题要求在大规模序列A中查找小规模序列B首次完全出现的位置。通过实现KMP算法，可以高效地解决这一挑战。

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 29660 Accepted Submission(s): 12477

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

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

题意：输入两个数组A[]和B[]的长度n，m和数组A，B，问数组B在数组A中首次出现的次数，即使得A[k]=B[0]，A[k+1]=B[1]...A[k+m-1]=B[m-1]的最小k，不存在的话就输出-1。

解题思路：其实就是字符串匹配的问题，裸的KMP，所以直接套用KMP算法就好啦。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN = 1e6 + 10; 
const int MAXM = 1e4 + 10;
int A[MAXN];//母串 
int B[MAXM];//子串 
int fail[MAXM];//失配数组 
int N,M;//数组长度 
void getFail()//得到失配函数 
{
	fail[0]=0;
	fail[1]=0;
	for(int i=1;i<M;i++)
	{
		int j=fail[i];
		while(j && B[i]!=B[j]) j=fail[j];
		fail[i+1]=B[i]==B[j]?j+1:0;
	}
}
int find()//查询函数 
{
	memset(fail,0,sizeof(fail));
	getFail();//得到失配数组 
	int j=0;
	for(int i=0;i<N;i++)//匹配 
	{
		while(j && B[j]!=A[i]) j=fail[j];
		if(A[i]==B[j]) j++;
		if(j==M) return i+2-M;//找到匹配的地方直接返回 
	}
	return -1;//不存在，返回-1 
}
int main(void)
{
	int T;//测试组数 
	scanf("%d",&T);
	while(T--)
	{
		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]);
		}
		printf("%d\n",find());
	}
	return 0;
}