Submission #599078

#	Time	Username	Problem	Language	Result	Execution time	Memory
599078		MohammadAghil	The short shank; Redemption (BOI21_prison)	C++17	80 / 100	306 ms	99872 KiB

prison

      #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 = 1e6 + 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], st[maxn], en[maxn], tt, inv[maxn];
bool vis[maxn];

void dfs(int r){
     inv[tt] = r, st[r] = tt++, vis[r] = true;
     for(int c: adj[r]) h[c] = h[r] + 1, dfs(c);
     en[r] = tt;
}


pp seg[maxn<<2];
int laz[maxn<<2];

void updNode(int x, int k){
     seg[x].ff += k, laz[x] += k;
}
void shift(int x){
     if(laz[x]){
          updNode(x<<1, laz[x]);
          updNode(x<<1|1, laz[x]);
          laz[x] = 0;
     }
}
pp get(){
     return seg[1];
}
int get(int i, int x = 1, int lx = 0, int rx = tt){
     if(lx + 1 == rx) return seg[x].ff;
     int mid = (lx + rx)>>1;
     shift(x);
     if(i < mid) return get(i, x<<1, lx, mid);
     return get(i, x<<1|1, mid, rx);
}
void upd(int l, int r, int k, int x = 1, int lx = 0, int rx = tt){
     if(l <= lx && r >= rx){
          updNode(x, k);
          return;
     }
     shift(x);
     int mid = (lx + rx)>>1;
     if(l < mid) upd(l, r, k, x<<1, lx, mid);
     if(mid < r) upd(l, r, k, x<<1|1, mid, rx);
     seg[x] = max(seg[x<<1], seg[x<<1|1]);
}
void build(int x = 1, int lx = 0, int rx = tt){
     if(lx + 1 == rx){
          seg[x] = {h[inv[lx]], lx};
          return;
     }
     int mid = (lx + rx)>>1;
     build(x<<1, lx, mid);
     build(x<<1|1, mid, rx);
     seg[x] = max(seg[x<<1], seg[x<<1|1]);
}

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);
               // er(i, par[i]);
          }else{
               while(sz(stk) && stk.back() <= e[i]) stk.pop_back();
          }
     }

     per(i,n-1,0) if(is[i]){
          if(!vis[i]) h[i] = 1, dfs(i);
          // er(i, h[i], st[i]);
     }
     build();

     // rep(i,0,tt) er(i, get(i));
     while(d--){
          auto[hh, id] = get(); 
          if(hh <= 0) break;      
          // er(inv[id]);
          int r = inv[id], lst = -1;
          while(r + 1 && vis[r]){
               int H = get(st[r]);
               ans++;
               // er(r, H);
               upd(st[r], en[r], -H);
               if(lst + 1) upd(st[lst], en[lst], H);
               lst = r, vis[r] = false, r = par[r];
          }
          // rep(i,0,tt) er(i, get(i));
     }

     cout << n - ans << endl;

     return 0; 
}

• Subtask 1: 15 / 15
• Subtask 2: 10 / 10
• Subtask 3: 20 / 20
• Subtask 4: 10 / 10
• Subtask 5: 25 / 25
• Subtask 6: 0 / 20