CF1225B2 TV Subscriptions (Hard Version)

题目描述

The only difference between easy and hard versions is constraints.

The BerTV channel every day broadcasts one episode of one of the kk TV shows. You know the schedule for the next nn days: a sequence of integers a_1, a_2, \dots, a_na1,a2,…,a**n ( 1 \le a_i \le k1≤a**i≤k ), where a_ia**i is the show, the episode of which will be shown in ii -th day.

The subscription to the show is bought for the entire show (i.e. for all its episodes), for each show the subscription is bought separately.

How many minimum subscriptions do you need to buy in order to have the opportunity to watch episodes of purchased shows dd ( 1 \le d \le n1≤d≤n ) days in a row? In other words, you want to buy the minimum number of TV shows so that there is some segment of dd consecutive days in which all episodes belong to the purchased shows.

输入格式

The first line contains an integer tt ( 1 \le t \le 100001≤t≤10000 ) — the number of test cases in the input. Then tt test case descriptions follow.

The first line of each test case contains three integers n, kn,k and dd ( 1 \le n \le 2\cdot10^51≤n≤2⋅105 , 1 \le k \le 10^61≤k≤106 , 1 \le d \le n1≤d≤n ). The second line contains nn integers a_1, a_2, \dots, a_na1,a2,…,a**n ( 1 \le a_i \le k1≤a**i≤k ), where a_ia**i is the show that is broadcasted on the ii -th day.

It is guaranteed that the sum of the values of nn for all test cases in the input does not exceed 2\cdot10^52⋅105 .

输出格式

Print tt integers — the answers to the test cases in the input in the order they follow. The answer to a test case is the minimum number of TV shows for which you need to purchase a subscription so that you can watch episodes of the purchased TV shows on BerTV for dd consecutive days. Please note that it is permissible that you will be able to watch more than dd days in a row.

输入输出样例

输入 #1复制

输出 #1复制

说明/提示

In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 11 and on show 22 . So the answer is two.

In the second test case, you can buy a subscription to any show because for each show you can find a segment of three consecutive days, consisting only of episodes of this show.

In the third test case in the unique segment of four days, you have four different shows, so you need to buy a subscription to all these four shows.

In the fourth test case, you can buy subscriptions to shows 3,5,7,8,93,5,7,8,9 , and you will be able to watch shows for the last eight days.

题解：

滑动窗口题目的数据加强版（对比\(Easy\,\,\, Version\)）

其实最初认识滑动窗口这种题的时候是学单调队列的时候。但是显然这道题不能用数据结构来做。那么我们回归滑动窗口的本质来尝试着做这道题。

滑动窗口的本质是啥？滑动啊！

那我们就可以进行模拟：先手动跑第一个线段（初始窗口），统计颜色出现的次数（存到\(cnt[i]\)数组，注意这个数组一定要开成\(10^6\).如果这个颜色是第一次出现（即\(cnt[i]=0\)），那么就把颜色数（\(sum\)变量）++）

然后开滑。每到一个新元素，先删除滑动之前最古老的那个元素。然后再把最新的元素加进来，然后看一看这个\(cnt[i]\)是否需要把\(sum\)也随着改动。然后就可以AC了。

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=2*1e5+10;
const int maxk=1e6+10;
int n,k,d;
int a[maxn];
int cnt[maxk],sum,ans;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(cnt,0,sizeof(cnt));
        sum=0;
        scanf("%d%d%d",&n,&k,&d);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        for(int i=1;i<=d;i++)
        {
            if(!cnt[a[i]])
                sum++;
            cnt[a[i]]++;
        }
        ans=sum;
        for(int i=d+1;i<=n;i++)
        {
            cnt[a[i-d]]--;
            if(!cnt[a[i-d]])
                sum--;
            if(!cnt[a[i]])
                sum++;
            cnt[a[i]]++;
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}