제출 #321543

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
321543		ryansee	Hedgehog Daniyar and Algorithms (IZhO19_sortbooks)	C++14	17 / 100	3095 ms	145544 KiB

sortbooks

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

#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define f first
#define s second
#define cbr cerr<<"hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x,y) (lower_bound(all(x),y)-x.begin())
#define ubd(x,y) (upper_bound(all(x),y)-x.begin())
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline long long rand(long long x, long long y) { return rng() % (y+1-x) + x; } //inclusivesss
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }

using ll=int; 
using ld=long double;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
using pi=pair<ll,ll>; using spi=pair<ll,pi>; using dpi=pair<pi,pi>; 

#define LLINF INF //((long long)1e18)
#define INF int(1e9+1e6)
#define MAXN (1000006)
struct node {
	int v,add,base;
	node*l,*r;
	node(int S,int E){
		int s=S,e=E,m=(s+e)>>1;
		v=-LLINF, add=-LLINF, base=0;
		if(s^e)l=new node(s,m),r=new node(m+1,e);
	}
	void update(int s,int e,int x,int y,int nval) {
		value(s,e);
		int m=(s+e)>>1;
		if(s==x&&e==y) {
			add = max(add, nval);
			return;
		}
		if(x>m) r->update(m+1,e,x,y,nval);
		else if(y<=m) l->update(s,m,x,y,nval);
		else l->update(s,m,x,m,nval),r->update(m+1,e,m+1,y,nval);
		v = max(l->value(s,m), r->value(m+1,e));
		base = max(l->base, r->base);
	}
	ll rmq(int s,int e,int x,int y) {
		if(x>y) return -LLINF;
		value(s,e);
		int m=(s+e)>>1;
		if(s==x&&e==y) return v;
		if(x>m) return r->rmq(m+1,e,x,y);
		else if(y<=m) return l->rmq(s,m,x,y);
		else return max(l->rmq(s,m,x,m), r->rmq(m+1,e,m+1,y));
	}
	void set(int s,int e,int x,int nval) {
		value(s,e);
		if(s==e) {
			v = nval + base;
			return;
		}
		int m=(s+e)>>1;
		if(x>m) r->set(m+1,e,x,nval);
		else l->set(s,m,x,nval);
		v = max(l->value(s,m), r->value(m+1,e));
		base = max(l->base, r->base);
	}
	ll value(int s,int e) {
		v = max(v, add + base);
		if(s^e) {
			l->add=max(l->add, add);
			r->add=max(r->add, add);
		}
		add=-LLINF;
		return v;
	}
	void set_base(int s,int e,int x,int nval) {
		value(s,e);
		if(s==e) {
			v-= base, base = nval, v+=base;
			return;
		}
		int m=(s+e)>>1;
		if(x>m) r->set_base(m+1, e, x, nval);
		else l->set_base(s, m, x, nval);
		v = max(l->value(s,m), r->value(m+1,e));
		base = max(l->base, r->base);
	}
} *seg;
int n,m,A[MAXN];
bool can[MAXN];
vector<int> v;
vector<pi> ans;
vector<tuple<int,int,int>> Q[MAXN];
#define gc getchar_unlocked
ll in(){
	ll x=0;
	char ch=gc();
	while(ch<'0'||ch>'9') ch=gc();
	while(ch>='0'&&ch<='9') {
		x=(x<<3)+(x<<1)+(ch&15);
		ch=gc();
	}
	return x;
}
int main() {
	n=in(),m=in();
	FOR(i,1,n)A[i]=in();
	seg=new node(1, n);
	FOR(i,1,m) {
		ll l,r,k;
		l=in(), r=in(), k=in();
		Q[r].pb({l, k, i});
	}
	auto add_ans=[&](ll ind,ll val) {
		// while(ans.size()&&ans.back().s<=val)ans.pop_back();
		ans.eb(ind,val);
	};
	auto add=[&](ll x){
		while(v.size() && A[v.back()] <= A[x]) {
			add_ans(v.back(), seg->rmq(1, n, siz(v), siz(v)));
			seg->set(1, n, siz(v), -LLINF);
			v.pop_back();
		}
		v.eb(x);
		seg->set_base(1, n, siz(v), A[x]);
	};
	auto upd=[&](ll x){
		ll st=-1, en=siz(v);
		while(en-st>1){// highest ind that is still > x
			ll mid=(st+en)>>1;
			if(A[v[mid]] > x) st=mid;
			else en=mid;
		}
		if(st<0) return;
		seg->update(1, n, 1, st+1, x);
	};
	FOR(r,1,n) {
		upd(A[r]);
		add(r);
		for(auto q:Q[r]) {
			ll l=get<0>(q);
			ll ind = lbd(v, l) + 1;
			// ll ind2 = lbd(ans, pi(l,-LLINF));
			ll hmm = 0;
			for(auto i:ans) if(i.f>=l) hmm=max(hmm,i.s);
			ll mx = max(hmm, seg->rmq(1, n, ind, siz(v)));
			if(mx <= get<1>(q)) can[get<2>(q)]=1;
		}
	}
	FOR(i,1,m) cout<<can[i]<<'\n';
}

• Subtask 1: 8 / 8
• Subtask 2: 9 / 9
• Subtask 3: 0 / 13
• Subtask 4: 0 / 17
• Subtask 5: 0 / 30
• Subtask 6: 0 / 23