제출 #599113

#제출 시각아이디문제언어결과실행 시간메모리
599113MohammadAghilThe short shank; Redemption (BOI21_prison)C++17
100 / 100
411 ms324904 KiB
#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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