Submission #213314

#	Time	Username	Problem	Language	Result	Execution time	Memory
213314		ryansee	Constellation 3 (JOI20_constellation3)	C++14	100 / 100	948 ms	62200 KiB

constellation3

#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 ph push
#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)
#define ubd(x, y) upper_bound(all(x), y)
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); }

typedef long long ll; 
typedef long double ld;
#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)
typedef pair<ll,ll>pi; typedef pair<ll,pi>spi; typedef pair<pi,pi>dpi;

#define LLINF ((long long)1e18)
#define INF int(1e9+1e6)
#define MAXN (200006)
ll n,m,A[MAXN];
vector<pi> S[MAXN], row[MAXN];
vector<int> reveal[MAXN];
int p[MAXN], mi[MAXN], mx[MAXN];
struct ufds_ {
	ufds_ () {
		FOR(i,0,MAXN-1) p[i]=mi[i]=mx[i]=i;
	}
	void merge(int x,int y){
		x=find(x),y=find(y);
		if(x==y)return;
		p[x]=y;
		mx[y]=max(mx[y],mx[x]);
		mi[y]=min(mi[y],mi[x]);
	}
	int find(int x) { return (p[x]==x) ? x : p[x]=find(p[x]); }
	bool same(int a,int b) { return find(a)==find(b); }
} ufds;
struct node {
	int s,e,m;
	ll v=0,add=0;
	node*l,*r;
	node(int S,int E){
		s=S,e=E,m=(s+e)>>1;
		v=add=0;
		if(s^e)l=new node(s,m),r=new node(m+1,e);
	}
	ll value(){
		v+=add;
		if(s^e){
			l->add+=add;
			r->add+=add;
		}
		add=0;
		return v;
	}
	void update(int x,int y,ll nval){
		if(x>y)return;
		value();
		if(s==x&&e==y){add+=nval;return;}
		if(x>m)r->update(x,y,nval);
		else if(y<=m)l->update(x,y,nval);
		else l->update(x,m,nval),r->update(m+1,y,nval);
		v=max(l->value(),r->value());
	}
	ll rmq(int x,int y){
		value();
		if(s==x&&e==y) return v;
		if(x>m) return r->rmq(x,y);
		else if(y<=m) return l->rmq(x,y);
		else return max(l->rmq(x,m),r->rmq(m+1,y));
	}
}*seg;
ll total;
int main(){
	FAST
	cin>>n;
	FOR(i,1,n)cin>>A[i], reveal[A[i]+1].pb(i);
	cin>>m;
	FOR(i,1,m){
		ll x,y,c;cin>>x>>y>>c;total+=c;
		S[x].eb(y,c);
	}
	FOR(i,1,n){
		sort(all(S[i]));
		vector<pi> tmp;
		FOR(j,0,siz(S[i])-1){
			if(tmp.empty() || tmp.back().s < S[i][j].s) tmp.eb(S[i][j]);
		}
		S[i]=tmp;
		DEC(j,siz(S[i])-1,1)S[i][j].s-=S[i][j-1].s;
		for(auto j:S[i]) row[j.f].eb(i, j.s);
	}
	seg=new node(1,n);
	FOR(i,1,n){
		for(auto j:reveal[i]) {
			if(j>1 && !ufds.same(j-1, j) && A[j-1]<=A[j]) { // j-1 and j
				ll mx1=seg->rmq(mi[ufds.find(j-1)], mx[ufds.find(j-1)]);
				ll mx2=seg->rmq(mi[ufds.find(j)],mx[ufds.find(j)]);
				seg->update(mi[ufds.find(j-1)],mx[ufds.find(j-1)],mx2);
				seg->update(mi[ufds.find(j)],mx[ufds.find(j)],mx1);
				ufds.merge(j-1,j);
			}
			if(j<n && !ufds.same(j+1, j) && A[j+1]<=A[j]) { // j+1 and j
				ll mx1=seg->rmq(mi[ufds.find(j+1)], mx[ufds.find(j+1)]);
				ll mx2=seg->rmq(mi[ufds.find(j)],mx[ufds.find(j)]);
				seg->update(mi[ufds.find(j+1)],mx[ufds.find(j+1)],mx2);
				seg->update(mi[ufds.find(j)],mx[ufds.find(j)],mx1);
				ufds.merge(j,j+1);
			}
		}
		for(auto j:row[i]){
			seg->update(j.f, j.f, j.s);
		}
		// cerr<<seg->rmq(1,n)<<'\n';
	}
	ll ans=0;
	FOR(i,1,n)if(ufds.find(i)==i){
		ans += seg->rmq(mi[i], mx[i]);
		// cerr<<seg->rmq(mi[i],mx[i])<<'\n';
	}
	cout<<total-ans<<'\n';
}

• Subtask 1: 14 / 14
• Subtask 2: 21 / 21
• Subtask 3: 65 / 65