链接:
https://codeforces.com/contest/1247/problem/B2
题意:
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.
思路:
双指针遍历。
代码:
#include <bits/stdc++.h>
typedef long long LL;
using namespace std;
const int MAXN = 2e5+10;
int Vis[MAXN*10];
int Day[MAXN];
int n, k, d;
int main()
{
ios::sync_with_stdio(false);
int t;
scanf("%d", &t);
while (t--)
{
scanf("%d%d%d", &n, &k, &d);
for (int i = 1;i <= n;i++)
scanf("%d", &Day[i]);
for (int i = 1;i <= n;i++)
Vis[Day[i]] = 0;
int res = k, tmp = 0;
for (int i = 1;i <= d;i++)
{
if (Vis[Day[i]] == 0)
tmp++;
Vis[Day[i]]++;
}
res = min(res, tmp);
for (int i = d+1;i <= n;i++)
{
if (Vis[Day[i-d]] == 1)
tmp--;
Vis[Day[i-d]]--;
if (Vis[Day[i]] == 0)
tmp++;
Vis[Day[i]]++;
res = min(res, tmp);
}
printf("%d
", res);
}
return 0;
}