Submission #823695

#	Time	Username	Problem	Language	Result	Execution time	Memory
823695		GusterGoose27	Comparing Plants (IOI20_plants)	C++17	0 / 100	12 ms	17020 KiB

plants

#pragma GCC optimize("Ofast")

#include "plants.h"
#include <bits/stdc++.h>

using namespace std;

const int MAXN = 2e5+5;
const int SZ = 1<<18;
const int inf = 1e9;
int R[MAXN];
int n;

typedef pair<int, int> pii;

pii operator+(pii a, pii b) {
	return pii(a.first+b.first, a.second+b.second);
}

class stree {
public:
	int mn[2*SZ];
	int lz[2*SZ];
	stree() {}
	void reset(bool tp) {
		memset(lz, 0, sizeof(int)*2*SZ);
		for (int i = SZ; i < 2*SZ; i++) {
			mn[i] = (tp && i < SZ+n) ? R[i-SZ] : inf;
		}
		for (int i = SZ-1; i > 0; i--) mn[i] = min(mn[2*i], mn[2*i+1]);
	}
	void activate(int p, int cur = 1, int lp = 0, int rp = SZ-1) {
		if (lp > p || rp < p) return;
		if (lp == rp) {
			mn[cur] = lz[cur];
			return;
		}
		int md = (lp+rp)/2;
		activate(p, 2*cur, lp, md);
		activate(p, 2*cur+1, md+1, rp);
		mn[cur] = min(mn[2*cur], mn[2*cur+1])+lz[cur];
	}
	void set_lz(int cur, int p) {
		lz[cur] += p;
		mn[cur] += p;
	}
	int find(int cur = 1, int lp = 0, int rp = SZ-1, int ulz = 0) { // finds a position that is 0
		if (mn[cur]+ulz > 0) return -1;
		if (lp == rp) {
			mn[cur] = inf;
			return lp;
		}
		ulz += lz[cur];
		int md = (lp+rp)/2;
		int lq = find(2*cur, lp, md, ulz);
		if (lq >= 0) {
			mn[cur] = min(mn[2*cur], mn[2*cur+1])+lz[cur];
			return lq;
		}
		int v = find(2*cur+1, md+1, rp, ulz);
		assert(v >= 0);
		mn[cur] = min(mn[2*cur], mn[2*cur+1])+lz[cur];
		return v;
	}
	void upd(int lv, int rv, int v, int cur = 1, int lp = 0, int rp = SZ-1) {
		if (lp > rv || rp < lv) return;
		if (lp >= lv && rp <= rv) {
			set_lz(cur, v);
			return;
		}
		int md = (lp+rp)/2;
		upd(lv, rv, v, 2*cur, lp, md);
		upd(lv, rv, v, 2*cur+1, md+1, rp);
		mn[cur] = min(mn[2*cur], mn[2*cur+1])+lz[cur];
	}
	void del(int p, int cur = 1, int lp = 0, int rp = SZ-1) { // can be absorbed into find
		if (lp > p || rp < p) return;
		if (lp == rp) {
			mn[cur] = inf;
			return;
		}
		int md = (lp+rp)/2;
		del(p, 2*cur, lp, md); del(p, 2*cur+1, md+1, rp);
		mn[cur] = min(mn[2*cur], mn[2*cur+1])+lz[cur];
	}
	int rt_zero(int lv, int cur = 1, int lp = 0, int rp = SZ-1, int ulz = 0) {
		if (rp < lv || mn[cur]+ulz > 0) return n;
		if (lp == rp) {
			return lp;
		}
		ulz += lz[cur];
		int md = (lp+rp)/2;
		int lq = rt_zero(lv, 2*cur, lp, md, ulz);
		if (lq < n) return lq;
		return rt_zero(lv, 2*cur+1, md+1, rp, ulz);
	}
};

stree *tree[2];
int k;

int before[MAXN]; // 0 => 1, 1 => 0, 2 => -1

// if strict, we don't allow x. Otherwise, we optimize for x

void make_strict() { // could x be taller than y
	tree[0]->reset(0);
	tree[1]->reset(1);
	for (int i = 0; i < n; i++) {
		int f1;
		while ((f1 = tree[1]->find()) >= 0) {
			tree[0]->activate(f1);
			tree[0]->upd(f1+1, min(n-1, f1+k-1), 1);
			if (f1+k-1 >= n) tree[0]->upd(0, f1+k-1-n, 1);
		}
		int v = tree[0]->find();
		if (v == -1) return;
		if (v == 0) continue;
		before[v]++;
		tree[0]->upd(v+1, min(n-1, v+k-1), -1);
		tree[0]->upd(0, v+k-1-n, -1);
		tree[1]->upd(max(0, v-k+1), v-1, -1);
		if (v-k+1 < 0) tree[1]->upd(n+v-k+1, n-1, -1);
	}
	assert(false);
}

void target(int cur = 0) { // we are trying to activate cur.
// because it is possible, we will never have a cycle
	before[cur]++;
	while (1) {
		int f1;
		while ((f1 = tree[1]->rt_zero(max(0, cur-k+1))) < cur) target(f1);
		if (cur-k+1 < 0) while ((f1 = tree[1]->rt_zero(n+cur-k+1)) < n) target(f1);
		int u = tree[1]->rt_zero(cur);
		if (u == n) u = tree[1]->rt_zero(0);
		if (u == cur) {
			tree[1]->upd(max(0, cur-k+1), cur-1, -1);
			if (cur-k+1 < 0) tree[1]->upd(n+cur-k+1, n-1, -1);
			tree[1]->del(cur);
			return;
		}
		else target(u);
	}
	assert(false);
}

void init(int K, vector<int> r) {
	k = K;
	n = r.size();
	assert(n <= 300);
	for (int i = 0; i < n; i++) {
		R[i] = r[i];
	}
	tree[0] = new stree(); // 0s to the left
	tree[1] = new stree(); // rval
	make_strict();
	tree[0]->reset(0);
	tree[1]->reset(1);
}

int compare_plants(int x, int y) {
	return 1-before[y];
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 14
• Subtask 3: 0 / 13
• Subtask 4: 0 / 17
• Subtask 5: 0 / 11
• Subtask 6: 0 / 15
• Subtask 7: 0 / 25