2016 ccpc(长春) Sequence I (kmp)

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原创最新推荐文章于 2026-10-14 11:29:18 发布 468 阅读 AI 写同款· GEO 优化 ›

2021 CCPC训练 F. 地图压缩（ kmp +字符Hash ) F. 地图压缩样例题意给一个二维字符矩阵，每次询问给定一个区间，求这个区间的最小循环节。思路首先对于一维字符串的最小循环节（无需完全覆盖），方法就是求出kmp的nxt数组， ans = len - nxt[len] 。对于二维的情况，容易发现行与列之间是不冲突的。于是我们对行求一遍哈希，对列求一遍哈希。询问时把区间的每一行的哈希值取出来做一遍kmp的预处理 ,再把每一列的哈希值取出来做一遍kmp的预处理，两个结果相乘就是答案。这样处理的复杂度是 O（阅读详情

Sequence I

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 958 Accepted Submission(s): 370

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

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当时暴力过的，没想过kmp，太菜了
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define maxx 1123456
int ne[1123456];
int s[maxx],t[maxx];
int p,sum;
void get_next(int l)
{
    int i=0,j=-1;
    ne[0]=-1;
    while(i<l)
    {
        if(j==-1 || t[i]==t[j])
        {
            i++;j++;ne[i]=j;
        }
        else
        {
            j=ne[j];
        }
    }
}
void kmp(int x,int ls,int lt)
{
    int i=x;
    int j=0;
    sum=0;
    while(i<ls)
    {
        if(j==-1 || s[i]==t[j])
        {
            i+=p;j++;
        }
        else
        {
            j=ne[j];
        }
        if(j==lt)
        {
            sum++;
            j=ne[j];
        }
    }

}
int main()
{
    int n,m,i;
    int ll;
    int tt;
    scanf("%d",&tt);
    for(ll=1;ll<=tt;ll++)
    {
        sum=0;
        scanf("%d%d%d",&n,&m,&p);
        for(i=0;i<n;i++)
        {
            scanf("%d",&s[i]);
        }
        for(i=0;i<m;i++)
        {
            scanf("%d",&t[i]);
        }
        get_next(m);
        int ans=0;
        for(i=1;i<=n;i++)
        {
            printf("%d ",ne[i]);
        }
        printf("\n");
        for(i=0;i<n&&i<p;i++)  //如果i大于p，就重复的找了
        {
            kmp(i,n,m);
            ans+=sum;
        }
        printf("Case #%d: %d\n",ll,ans);
    }

    return 0;
}