hdu 5918 Sequence I kmp

最新推荐文章于 2018-09-24 21:54:00 发布

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本文探讨了一个特定的序列匹配问题，即在给定间隔p的情况下，如何有效地寻找子序列在母序列中出现的所有位置。通过KMP算法的巧妙应用，并适当调整搜索步长，实现了高效的子序列查找。

Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 470 Accepted Submission(s): 184

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

Source

2016中国大学生程序设计竞赛（长春）-重现赛

母串和子串，求子串在母串中出现的次数，但是遍历母串的时候会有间隔p，题目说的很明白，即 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 。

q+(m−1)p≤n and q≥1

.

思路：修改母串匹配子串的时候，每次跳p位，但是要注意起点要选0->p-1这期间所有点，不然会有点不会走到，vis数组去一下重即可。

#include<cstdio>
#include<cstring>
#include <iostream>
using namespace std;
const int maxn=1000010;
int T[maxn],P[maxn];
  int Next[maxn];
  int vis[maxn];
  int n,m,p;
void makeNext(int  P[],int m,int Next[])
{
    int q,k;
    Next[0] = 0;
    for (q = 1,k = 0; q < m; ++q)
    {
        while(k > 0 && P[q] != P[k])
            k = Next[k-1];
        if (P[q] == P[k])
        {
            k++;
        }
        Next[q] = k;
    }
}

int kmp(int T[],int P[],int n,int m,int Next[],int p)
{
    int i,q;
    int cnt=0;
    makeNext(P,m,Next);
        for(int s=0;s<p;s++)
    for (i = s,q = 0; i < n; i+=p)
    {
        while(q > 0 && P[q] != T[i])
            q = Next[q-1];
        if (P[q] == T[i])
        {
            q++;
        }
        if (q == m)
        {
            if(vis[i]==0)
            {cnt++;
            vis[i]=1;}
        }
    }
    return cnt;
}

int main()
{
    int t;
    scanf("%d",&t);
    int cns=0;
    while(t--)
    {
        memset(T,-1,sizeof(T));
        memset(P,-2,sizeof(P));
        memset(vis,0,sizeof(vis));
        memset(Next,0,sizeof(Next));
        scanf("%d%d%d",&n,&m,&p);
        for(int i=0;i<n;i++)
            scanf("%d",&T[i]);
        for(int i=0;i<m;i++)
            scanf("%d",&P[i]);
        //cout<<kmp(T,P,n,m,Next,p)<<endl;
        printf("Case #%d: %d\n",++cns,kmp(T,P,n,m,Next,p));
    }
}