B1. TV Subscriptions (Easy Version)
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 kk TV shows. You know the schedule for the next nn days: a sequence of integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k), where aiai 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≤d≤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.
Input
The first line contains an integer tt (1≤t≤1001≤t≤100) — 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≤n≤1001≤n≤100, 1≤k≤1001≤k≤100, 1≤d≤n1≤d≤n). The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤k1≤ai≤k), where aiai 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 100100.
Output
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.
Example
input
Copy
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
Copy
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 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.
题意：给出一个数列 找出确定窗口长度为d
滑动时维护这个窗口即可
这里k的值可能很大
所以b2用vis数组会tle
#include<bits/stdc++.h>
using namespace std;
int vis[1000005];
int a[1000005];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
queue<int>q;
int n,k,d;
scanf("%d%d%d",&n,&k,&d);
for(int i=1;i<=k;i++) vis[i]=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int pos=0;
int ans=1e9;
for(int i=1;i<=d;i++)
{
vis[a[i]]++;
if(vis[a[i]]==1) pos++;
q.push(a[i]);
}
ans=pos;
for(int i=d+1;i<=n;i++)
{
vis[q.front()]--;
if(vis[q.front()]==0) pos--;
q.pop();
q.push(a[i]);
vis[a[i]]++;
if(vis[a[i]]==1) pos++;
ans=min(pos,ans);
}
printf("%d
",ans);
}
}