제출 #575795

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
575795		jiahng	The short shank; Redemption (BOI21_prison)	C++14	0 / 100	29 ms	62932 KiB

prison

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
//~ #define ll int
#define int ll
typedef pair<int32_t, int32_t> pi;
typedef vector <int> vi;
typedef vector <pi> vpi;
typedef pair<pi, ll> pii;
typedef set <ll> si;
typedef long double ld;
#define f first
#define s second
#define mp make_pair
#define FOR(i,s,e) for(int i=s;i<=int(e);++i)
#define DEC(i,s,e) for(int i=s;i>=int(e);--i)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define maxn 2000001
#define INF 1e9
#define MOD 1000000007
typedef pair <vi, int> pvi;
typedef pair <int,pi> ipi;
typedef vector <pii> vpii;

int N,D,T,A[maxn],L[maxn];
int pre[maxn]; // closest which covers it
bool spec[maxn];
vi adj[maxn];

struct node{
	int s,e,m,lazy=0;
	pi val = pi(0,-1);
	node *l,*r;
	
	node (int ss,int ee){
		s = ss; e = ee; m = (s + e) / 2;
		val.s = s;
		if (s != e){
			l = new node(s,m);
			r = new node(m+1,e);
		}
	}
	void prop(){
		if (s == e || lazy == 0) return;
		l->val.f += lazy; r->val.f += lazy;
		l->lazy += lazy; r->lazy += lazy;
		lazy = 0;
	}
	pi qry(int a,int b){
		prop();
		if (a <= s && e <= b) return val;
		else if (a > e || s > b) return pi(-1,-1);
		return max(l->qry(a,b), r->qry(a,b));
	}
	void upd(int a,int b,int c){
		prop();
		if (a <= s && e <= b){
			val.f += c;
			lazy += c;
		}else if (a > e || s > b) return;
		else{
			l->upd(a,b,c); r->upd(a,b,c);
			val = max(l->val, r->val);
		}
	}
}*root;

int st[maxn], en[maxn],unst[maxn],depth[maxn], co = 1;
void dfs(int x){
	unst[co] = x;
	st[x] = co++;
	aFOR(i,adj[x]){
		depth[i] = depth[x] + 1; dfs(i);
	}
	en[x] = co - 1;
}
int nxt[maxn]; bool del[maxn];
pi mxdepth[maxn];

int32_t main(){
    fast;
    cin >> N >> D >> T;
    FOR(i,1,N) cin >> A[i];
    stack <pi> sta;
    // j covers i if A[j] + i - j <= T
    // A[j] + j <= T - i
    // increasing stack
    
    FOR(i,1,N){
		while (!sta.empty() && sta.top().f >= A[i] - i) sta.pop();
		sta.push(pi(A[i] - i, i));
		while (!sta.empty() && sta.top().f > T - i) sta.pop();
		if (!sta.empty()) pre[i] = sta.top().s;
	}

    int mn = INF;
    int ans = 0; //number which will rebel (covered)
    
    FOR(i,1,N){
		mn = min(mn, A[i] - i);
		if (A[i] > T && mn <= T - i) spec[i] = 1;
		
		if (mn <= T - i) ans++; // will rebel if no barriers
	}
	
	int r = 1;
	FOR(i,1,N) if (spec[i]){
		r = max(r, i+1);
		while (r <= N && (!spec[r] || pre[r] > i)) r++;
		nxt[i] = r;
		if (r <= N){ // r is parent
			adj[r].pb(i);
			//~ cout << r << ' ' << i << '\n';
		}
	}
	//~ cout << spec[3] << ' ' << nxt[3] << '\n';
	//~ return 0;
	mem(st, -1);
	DEC(i,N,1) if (spec[i] && st[i] == -1){
		depth[i] = 1; dfs(i);
	}
	
	DEC(i,N,1){
		mxdepth[i] = pi(depth[i], i);
		aFOR(j,adj[i]) mxdepth[i] = max(mxdepth[i], mxdepth[j]);
	}
	set <pi> roots;
	FOR(i,1,N) if (nxt[i] > N && spec[i]) roots.insert(pi(mxdepth[i].f, i));
	
	FOR(i,1,D){
		if (roots.empty()) break;
		int optRoot = roots.rbegin()->s; roots.erase(--roots.end());
		int opt = mxdepth[optRoot].s;
		
		ans -= mxdepth[optRoot].f;
		
		for (int x = opt; x <= N && !del[x]; x = nxt[x]){
			del[x] = 1;
			aFOR(j, adj[x]) if (!del[j]) roots.insert(pi(mxdepth[j].f, j));
		}
	}
	cout << ans;
		
	
	
}

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