Submission #227295

#	Time	Username	Problem	Language	Result	Execution time	Memory
227295		ryansee	Magic Tree (CEOI19_magictree)	C++14	0 / 100	75 ms	24800 KiB

magictree

#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); }

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 (100006)
ll n, m, k, old[MAXN];
vector<int> v[MAXN];
pi A[MAXN];
set<int> s[MAXN];
struct node{
	ll prior, sz, key, val, lazy, mx;
	node*l=0,*r=0;
};
typedef node* pnode;
vector<pnode> treap;
pnode init(ll _key,ll nval){
	pnode tmp=new node;
	tmp->prior=rng(), tmp->sz=1, tmp->key=_key, tmp->val=nval, tmp->lazy=0, tmp->mx=nval;
	tmp->l=0, tmp->r=0;
	return tmp;
}
void push(pnode& t){
	if(!t)return;
	t->val += t->lazy, t->mx += t->lazy;
	if(t->l) t->l->lazy+=t->lazy;
	if(t->r) t->r->lazy+=t->lazy;
	t->lazy=0;
}
void pull(pnode& t){
	if(!t)return;
	if(t->l) push(t->l);
	if(t->r) push(t->r);
	t->mx=max({(t->l?t->l->mx:0),(t->r?t->r->mx:0),t->val});
}
void upd(pnode& t){
	if(!t) return;
	t->sz = (t->l?t->l->sz:0) + (t->r?t->r->sz:0) + 1;
}
void split(pnode t,pnode&l,pnode&r,ll nval){
	if(!t) return void(l=r=0);
	push(t);
	if(t->key <= nval) split(t->r, t->r, r, nval), l=t;
	else split(t->l,l,t->l,nval),r=t;
	upd(t), pull(t);
}
void unite(pnode&t,pnode l,pnode r){
	if(!l || !r) return void(t=l?l:r);
	push(l),push(r);
	if(l->prior < r->prior) swap(l,r);
	pnode L,R;
	split(r,L,R,l->key);
	unite(l->l,l->l,L);
	unite(l->r,l->r,R);
	t=l;
	upd(t);
	pull(t);
}
bool find(pnode&t,ll key){
	if(!t) return 0;
	if(t->key==key) return 1;
	else if(t->key>key) return find(t->l,key);
	else return find(t->r,key);
}
void update(pnode&t,ll l,ll r,ll nval){
	pnode L,R;
	split(t,L,t,l-1);
	split(t,t,R,r);
	if(t) t->lazy += nval;
	unite(t,t,R);
	unite(t,L,t);
}
ll rmq(pnode&t,ll l,ll r){
	pnode L,R;
	split(t,L,t,l-1);
	split(t,t,R,r);
	ll ans=t ? t->mx+t->lazy : 0;
	unite(t,t,R);
	unite(t,L,t);
	return ans;
}
void dfs(int x,int p){
	old[x]=-1;
	for(auto i:v[x]) if(i^p) {
		dfs(i, x);
		if(old[x]==-1 || s[old[i]].size()>s[old[x]].size()) old[x]=old[i];
	}
	if(old[x]==-1){
		old[x]=treap.size();
		treap.eb(pnode());
		treap.back()=0;
	}
	// i is updated by j, where j is largest < i
	for(auto i:v[x]) if((i^p) && (old[i]^old[x])) {
		for(auto j=prev(s[old[i]].end());;--j){
			ll pre=rmq(treap[old[x]],0,*j);
			ll myval=rmq(treap[old[i]],*j,*j);
			if(!find(treap[old[x]],*j))unite(treap[old[x]],treap[old[x]],init(*j,myval+pre));
			else update(treap[old[x]],*j,*j,myval);
			update(treap[old[x]],(*j)+1,INF,myval);
			s[old[x]].ins(*j);
			if(j==s[old[i]].begin()) break;
		}
	}
	if(A[x].f){
		ll pre=rmq(treap[old[x]],0,A[x].f);
		if(!find(treap[old[x]],A[x].f))unite(treap[old[x]],treap[old[x]],init(A[x].f,A[x].s+pre));
		else update(treap[old[x]],A[x].f,A[x].f,A[x].s);
		s[old[x]].ins(A[x].f);
	}
	if(x==1){
		cout<<rmq(treap[old[x]],0,INF)<<'\n';
	}
}
int main(){
	FAST
	cin>>n>>m>>k;
	FOR(i,2,n){
		ll p;cin>>p;
		v[p].eb(i), v[i].eb(p);
	}
	FOR(i,1,m){
		ll a;cin>>a;
		cin>>A[a].f>>A[a].s;
		assert(A[a].f);
	}
	dfs(1,1);
}

• Subtask 1: 0 / 6
• Subtask 2: 0 / 3
• Subtask 3: 0 / 11
• Subtask 4: 0 / 12
• Subtask 5: 0 / 16
• Subtask 6: 0 / 13
• Subtask 7: 0 / 22
• Subtask 8: 0 / 17