Submission #336662

#	Time	Username	Problem	Language	Result	Execution time	Memory
336662		syy	Candies (JOI18_candies)	C++17	100 / 100	167 ms	14836 KiB

candies

/* Transition from k to k+1 will always be
 * ...0101000... to ...0101010..., or
 * ...0101010... to ...1010101...
 * Thus, store the possible states that it can go in a pq
 */
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define FOR(i, a, b) for(ll i = (ll)a; i <= (ll)b; i++)
#define DEC(i, a, b) for(ll i = (ll)a; i >= (ll)b; i--)
typedef pair<ll, ll> pi;
typedef pair<pi, ll> pii;
typedef pair<ll, pi> ipi;
typedef pair<pi, pi> pipi;
#define f first
#define s second
typedef vector<ll> vi;
typedef vector<pi> vpi;
typedef vector<pii> vpii;
#define pb push_back
#define pf push_front
#define all(v) v.begin(), v.end()
#define size(v) (ll) v.size()
#define disc(v) sort(all(v)); v.resize(unique(all(v)) - v.begin());
#define INF (ll) 1e9 + 100
#define LLINF (ll) 1e18
#define fastio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define sandybridge __attribute__((optimize("Ofast"), target("arch=sandybridge")))
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline ll rand(ll x, ll y) { ++y; return (rng() % (y-x)) + x; } //inclusivesss

ll n, arr[200005], ss[200005], p[200005], L[200005], R[200005], ans;
bool vis[200005];
priority_queue<ipi> pq;

inline ll getans(ll x, ll y) {
	return (y % 2 ? ss[y] - ss[x-1] : ss[x-1] - ss[y]);
}

ll fs(ll x) {
	if (p[x] == x) return x;
	return p[x] = fs(p[x]);
}

int main() {
	fastio; cin >> n;
	FOR(i, 1, n) {
		cin >> arr[i];
		ss[i] = ss[i-1] + (i % 2 ? arr[i] : -arr[i]);
		p[i] = L[i] = R[i] = i;
		pq.push(ipi(arr[i], pi(i, i)));
	}
	FOR(i, 1, (n+1)/2) {
		ll l, r;
		while (true) {
			l = pq.top().s.f;
			r = pq.top().s.s;
			pq.pop();
			if (!vis[l-1] and !vis[r+1]) break;
		}
		ans += getans(l, r);
		cout << ans << "\n";
		vis[l] = 1;
		vis[r] = 1;
		// because at every step you expand at most 1
		// you only need to set vis of l and r as
		// all the intermediate nodes must have been set before
		ll par;
		if (l == r) par = fs(l);
		else {
			par = fs(l+1);
			p[fs(l)] = par;
			p[fs(r)] = par;
			L[par] = l, R[par] = r;
		}
		if (vis[r+2]) {
			R[par] = R[fs(r+2)];
			p[fs(r+2)] = par;
		}
		if (l > 1 and vis[l-2]) {
			L[par] = L[fs(l-2)];
			p[fs(l-2)] = par;
		}
		if (L[par] > 1 and R[par] < n) pq.push(ipi(getans(L[par]-1, R[par]+1), pi(L[par]-1, R[par]+1)));
	}
}

• Subtask 1: 8 / 8
• Subtask 2: 92 / 92