hdu 5918(KMP)

最新推荐文章于 2019-09-24 16:09:12 发布

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最新推荐文章于 2019-09-24 16:09:12 发布

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本文链接：https://blog.youkuaiyun.com/huyiqun753/article/details/52744063

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，通过将原数组拆分成子数组并分别与目标序列进行匹配，实现了高效查找。

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Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 533 Accepted Submission(s): 210

Problem Description

Mr. Frog has two sequences

a1,a2,⋯,an and

b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence

b1,b2,⋯,bm is exactly the sequence

aq,aq+p,aq+2p,⋯,aq+(m−1)p where

q+(m−1)p≤n and

q≥1.

Input

The first line contains only one integer

T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers

1≤n≤106,1≤m≤106 and

1≤p≤106.

The second line contains n integers

a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers

b1,b2,⋯,bm(1≤bi≤109).

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

Sample Input

2
6 3 1
1 2 3 1 2 3
1 2 3
6 3 2
1 3 2 2 3 1
1 2 3

Sample Output

Case #1: 2
Case #2: 1

题意：

给出两个数组a，b与p，求有多少个q满足aq,aq+p,aq+2p,⋯,aq+(m−1)p等于b数组。

思路：

把a数组分为p个数组，分别于b数组进行kmp匹配。

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string.h>
#include<vector>
#include<bitset>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<math.h>
#include<map>
using namespace std;
int gcd(int a,int b)
{
	if(b==0)
		return a;
	else return gcd(b,a%b);
}
const int N=2000005;
int a[N],b[N],s[N];
int f[N];
int n,m,p;
void getfill()  
{  
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)  
    {  
        int j=f[i];  
        while(j && b[i]!=b[j])  
            j=f[j];  
        f[i+1]=(b[i]==b[j])?j+1:0;  
    }  
}  
int find(int len)  
{  
    int ans=0;  
	int j=0;
    for(int i=0;i<len;i++)  
    {  
        while(j && s[i]!=b[j])  
            j=f[j];  
        if(s[i]==b[j])  
            j++;  
        if(j==m)
            ans++;
    }
    return ans;
}
int main()
{
	freopen("in.txt","r",stdin);
	freopen("out.txt","w",stdout);
	int t;
	scanf("%d",&t);
	int num=1;
	while(t--)
	{
		scanf("%d%d%d",&n,&m,&p);
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		for(int i=0;i<m;i++)
			scanf("%d",&b[i]);
		int sum=0;
		getfill();
		for(int i=0;i<p;i++)
		{
			int len=0;
			for(int j=i;j<n&&i+(m-1)*p<n;j+=p)
				s[len++]=a[j];
			sum+=find(len);
		}
		printf("Case #%d: %d\n",num,sum);
		num++;
	}
    return 0;
}