time limit per test：2 seconds
memory limit per test：256 megabytes
input：standard input
output：standard output
The only difference between easy and hard versions is constraints.
The BerTV channel every day broadcasts one episode of one of the k TV shows. You know the schedule for the next n days: a sequence of integers a1,a2,…,an (1≤ai≤k), where ai is the show, the episode of which will be shown in i-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 d (1≤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 d consecutive days in which all episodes belong to the purchased shows.
Input
The first line contains an integer t (1≤t≤10000) — the number of test cases in the input. Then t test case descriptions follow.
The first line of each test case contains three integers n,k and d (1≤n≤2⋅105, 1≤k≤106, 1≤d≤n). The second line contains n integers a1,a2,…,an (1≤ai≤k), where ai is the show that is broadcasted on the i-th day.
It is guaranteed that the sum of the values of n for all test cases in the input does not exceed 2⋅105.
Output
Print t 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 d consecutive days. Please note that it is permissible that you will be able to watch more than d days in a row.
Example
input
4
5 2 2
1 2 1 2 1
9 3 3
3 3 3 2 2 2 1 1 1
4 10 4
10 8 6 4
16 9 8
3 1 4 1 5 9 2 6 5 3 5 8 9 7 9 3
output
2
1
4
5
Note
In the first test case to have an opportunity to watch shows for two consecutive days, you need to buy a subscription on show 1 and on show 2. 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 3,5,7,8,9, and you will be able to watch shows for the last eight days.
思路：
剛開始看到這個題目時，想到了莫隊的分塊統計+埃篩，寫完了一直在debug，然後看了學長的代碼明白過來了，原來是這樣做
t組輸入樣例，
n表示n個數，k表示着n個數中的最大值，d表示連續的幾個數中不同的數字個數（不重複的數字個數）
我可以表示出第一個區間裏符合題意的樹，然後依次往下移動，知道所有的都找完
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
using namespace std;
const int maxn = 1e6 + 10;
set<int> s;
map<int,int> mp;
int a[maxn];
int t,n,k,d;
int main()
{
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cin >> t;
while(t--){
mp.clear();
s.clear();
cin >> n >> k >> d;
for(int i = 1; i <= n; i++)
cin >> a[i];
for(int i = 1; i <= d; i++){
mp[a[i]]++;
s.insert(a[i]);
}
//左右邊界找到，然後依次下移，利用的map的計數和set的去重
int l = 1, r = d,ans = s.size();
while(r < n){//等於不成立，因爲裏面有r++
mp[a[l]]--;//我是想要右移
if(mp[a[l]]==0)//沒有的話，就可以刪除
s.erase(a[l]);
l++;
r++;
s.insert(a[r]);//向右移動，插入
mp[a[r]]++;//這一步剛開時忘記寫，
if(s.size() < ans)
ans = s.size();
}
cout << ans << endl;
}
return 0;
}