이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
// #pragma GCC optimize ("Ofast,unroll-loops")
// #pragma GCC target ("avx2")
using namespace std;
typedef long long ll;
typedef pair<int, int> pp;
#define er(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << endl
#define per(i,r,l) for(int i = (r); i >= (l); i--)
#define rep(i,l,r) for(int i = (l); i < (r); i++)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define pb push_back
#define ss second
#define ff first
void err(istringstream *iss){}template<typename T,typename ...Args> void err(istringstream *iss,const T &_val, const Args&...args){string _name;*iss>>_name;if(_name.back()==',')_name.pop_back();cerr<<_name<<" = "<<_val<<", ",err(iss,args...);}
void IOS(){
cin.tie(0) -> sync_with_stdio(0);
// #ifndef ONLINE_JUDGE
// freopen("in.in", "r", stdin);
// freopen("out.out", "w", stdout);
// #endif
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll mod = 1e9 + 7, maxn = 2e6 + 5, lg = 22, inf = ll(1e9) + 5;
ll pw(ll a,ll b,ll md=mod){if(!b)return 1;ll k=pw(a,b>>1ll);return k*k%md*(b&1ll?a:1)%md;}
vector<int> adj[maxn];
int par[maxn], h[maxn];
pp ptr[maxn];
bool vis[maxn];
void dfs(int r){
vis[r] = true, ptr[r] = {1, r};
for(int c: adj[r]) h[c] = h[r] + 1, dfs(c), ptr[r] = max(ptr[r], {ptr[c].ff + 1, ptr[c].ss});
}
int main(){ IOS();
int n, d, t; cin >> n >> d >> t;
vector<int> a(n), e(n);
rep(i,0,n) cin >> a[i], e[i] = t - a[i] + i;
vector<bool> is(n);
int mx = -1, ans = 0;
rep(i,0,n){
if(a[i] > t) {
if(mx < i) ans++;
else is[i] = true;
}
mx = max(mx, e[i]);
}
vector<int> stk;
per(i,n-1,0){
if(is[i]){
par[i] = -1;
if(sz(stk)) par[i] = stk.back(), adj[par[i]].pb(i);
stk.pb(i);
}else{
while(sz(stk) && stk.back() <= e[i]) stk.pop_back();
}
}
vector<vector<int>> q(n + 1);
per(i,n-1,0) if(is[i]){
if(!vis[i]) h[i] = 1, dfs(i), q[ptr[i].ff].pb(i);
}
per(i,n,1){
for(int c: q[i]){
int lst = -1, r = ptr[c].ss;
while(r - par[c]){
ans++;
for(int t: adj[r]) if(t - lst){
q[ptr[t].ff].pb(t);
}
lst = r, r = par[r];
}
d--;
if(!d) break;
}
if(!d) break;
}
cout << n - ans << endl;
return 0;
}
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