CodeForces 367B Sereja ans Anagrams

最新推荐文章于 2021-06-26 15:19:56 发布

Davenny

最新推荐文章于 2021-06-26 15:19:56 发布

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本文介绍了一种用于寻找两个序列中特定位置匹配的算法。通过枚举起点并利用map数据结构进行元素频率比较，实现高效查找。适用于序列长度较大情况。

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Description
Sereja has two sequences a and b and number p. Sequence a consists of n integers a1, a2, …, an. Similarly, sequence b consists of m integers b1, b2, …, bm. As usual, Sereja studies the sequences he has. Today he wants to find the number of positions q(q + (m - 1)·p ≤ n; q ≥ 1), such that sequence b can be obtained from sequence aq, aq + p, aq + 2p, …, aq + (m - 1)p by rearranging elements.

Sereja needs to rush to the gym, so he asked to find all the described positions of q.

Input
The first line contains three integers n, m and p(1 ≤ n, m ≤ 2·105, 1 ≤ p ≤ 2·105). The next line contains n integers a1, a2, …, an(1 ≤ ai ≤ 109). The next line contains m integers b1, b2, …, bm(1 ≤ bi ≤ 109).

Output
In the first line print the number of valid qs. In the second line, print the valid values in the increasing order.

Sample Input
Input
5 3 1
1 2 3 2 1
1 2 3
Output
2
1 3
Input
6 3 2
1 3 2 2 3 1
1 2 3
Output
2
1 2

思路：
首先map可以直接用等号来判断两个map是否一样。
其次，我们枚举起点，而且外层只用从1枚举到p,内后像队列一样删掉对头，队尾插入来维护map，即可；

#include <cstdio>
#include <map>
#include <algorithm>
using namespace std;
int n,m,p;
int cnt=0,q;
int a[2000090];
map<int,int> tmp1,tmp2;
int ans[2000090];
int main()
{
    scanf("%d%d%d",&n,&m,&p);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    int t;
    for(int i=0;i<m;i++)
    {
        scanf("%d",&t);
        tmp1[t]++;
    }
    for(int i=1;i<=p;i++)
    {
        tmp2.clear();
        for(long long j=0;j<m&&i+j*p<=n;j++)
         tmp2[a[i+j*p]]++;
         if(tmp1==tmp2)
            ans[cnt++]=i;
         for(long long j=m;i+j*p<=n;j++)
         {
             int pre=a[i+(j-m)*p];
             if(tmp2[pre]==1)
                tmp2.erase(pre);
             else tmp2[pre]--;
             tmp2[a[i+j*p]]++;
             if(tmp2==tmp1)
                ans[cnt++]=i+(j-m+1)*p;
         }
    }
    printf("%d\n",cnt);
    sort(ans,ans+cnt);
    for(int i=0;i<cnt;i++)
    {
        if(i) printf(" ");
        printf("%d",ans[i]);
    }
    puts("");
    return 0;
}