Submission #1156058

#	Time	Username	Problem	Language	Result	Execution time	Memory
1156058		tesseract	Game (IOI13_game)	C++20	63 / 100	7388 ms	321536 KiB

game

#include "game.h"
#include <cassert>
#include <vector>
#include <map>

using namespace std;
long long gcd2(long long X, long long Y);
template<class Key, class T, class Compare, class Allocator>
T getOrDefault(const map<Key, T, Compare, Allocator> &m, const Key &k, const T &def) {
	auto it = m.find(k);
	return it == m.end() ? def : it->second;
}

// Intended for nonempty intervals.
struct Interval {
	int L, U;

	inline bool operator==(const Interval &other) const = default;

	inline int length() const {return U-L;}

	inline bool contains(const Interval &x) const { return L <= x.L && x.U <= U; }
	inline bool contains(int x) const { return L <= x && x < U;}
	inline bool nonempty() const { return length() > 0; }
	inline int mid() const { return L + (U-L)/2;}
	inline Interval leftHalf() const { return Interval {L, mid()};}
	inline Interval rightHalf() const { return Interval {mid(), U};}
};

// Intended for intervals known to intersect.
inline Interval intersection(const Interval &x, const Interval &y) {
	return {max(x.L, y.L), min(x.U, y.U)};
}


inline bool intersects(const Interval &x, const Interval &y) {
	return intersection(x, y).nonempty();
}


class LazySegTreeIntervals {
private:
	int nextId = 0;

	struct Node {
		int id;
		Interval ivl;
		Node *left, *right;

		Node(int id_, const Interval &ivl_) : id(id_), ivl(ivl_), left(nullptr), right(nullptr) {}
		inline int length() const { return ivl.length();}
	};

	Node *root;
	int N;

	inline Node *createNewNode(const Interval &ivl) {
		return new Node(nextId++, ivl);
	}

	void decompose(const Node *node, const Interval &queryIvl, vector<int> &ids) const {
		assert(node->ivl.contains(queryIvl));
		if(node->ivl == queryIvl) {
			ids.push_back(node->id);
			return;
		}
		if(node->left && intersects(node->left->ivl, queryIvl)) {
			decompose(node->left, intersection(node->left->ivl, queryIvl), ids);
		}
		if(node->right && intersects(node->right->ivl, queryIvl)) {
			decompose(node->right, intersection(node->right->ivl, queryIvl), ids);
		}
	}


public:

	struct ParentChildIds {
		int p, c1, c2;
	};
private:
	inline ParentChildIds getParentChildId(Node *node) {
		return {node->id, node->left ? node->left->id : NO_ID, node->right? node->right->id : NO_ID};
	}
public:


	inline LazySegTreeIntervals(int N_) : N(N_) {
		assert(N >= 1);
		root = createNewNode({0,N});
	}


	constexpr static int NO_ID = -1;

	// Ensure that all nodes on the path from the root [0,N) to the
	// leaf [x, x+1) exist. For each such node, returns a tuple
	// with the id of that node and both its children (or NO_ID if a child
	// doesn't exist). These are retured in order from root to leaf.
	vector<ParentChildIds> ensureNodesAndGetIds(int x) {
		assert(0 <= x && x < N);

		vector<ParentChildIds> ids;
		ids.reserve(32); // Upper bound for maximum height of the tree for the maximum N

		Node *cur = root;
		while(true) {
			assert(cur && cur->ivl.contains(x));
			if(cur->ivl.length() == 1) {
				ids.push_back(getParentChildId(cur));
				break;
			}
			int mid = cur->ivl.mid();
			if(x < mid) {
				if (!cur->left) cur->left = createNewNode(cur->ivl.leftHalf());
				ids.push_back(getParentChildId(cur));
				cur = cur->left;
			} else {
				if (!cur->right) cur->right = createNewNode(cur->ivl.rightHalf());
				ids.push_back(getParentChildId(cur));
				cur = cur->right;
			}
		}
		return ids;
	}

	// Try to decompose [a,b) into constituent intervals as
	// if this was a full segtree. Let X be the set of values for
	// which createIntervalsAndGetIds has been called.
	// Consider the ids returned by this method, and
	// let Y be the union of their corresponding intervals.
	// Then we have Y intersection X = [a,b) intersection X.
	// If this was a full segtree, we would have Y = [a,b).
	// Of course, we also have that the size of the returned vector
	// is at most logarithmic in N.
	vector<int> decomposeIntervalAndGetIds(const Interval &ivl) const {
		vector<int> ids;
		decompose(root, ivl, ids);
		return ids;
	}
};

LazySegTreeIntervals Rtree(1), Ctree(1);


// A key (x,y) corresponds to node with id x in Rtree and
// a node with id y in Ctree. Each of those nodes represent an
// interval of indices. Their product represents a subrectangle
// of cells. The value is the gcd of the values in those cells.
map<pair<int, int>, long long> gcds;

void init(int R, int C) {
	Rtree = LazySegTreeIntervals(R);
	Ctree = LazySegTreeIntervals(C);
}

void update(int P, int Q, long long K) {
	// Each of these is ordered from root to leaf.
	vector<LazySegTreeIntervals::ParentChildIds> rIds = Rtree.ensureNodesAndGetIds(P);
	vector<LazySegTreeIntervals::ParentChildIds> cIds = Ctree.ensureNodesAndGetIds(Q);

	assert(!rIds.empty());
	assert(!cIds.empty());

	for (auto rit = rIds.rbegin(); rit != rIds.rend(); ++rit)
	for (auto cit = cIds.rbegin(); cit != cIds.rend(); ++cit) {
		if (rit == rIds.rbegin() && cit == cIds.rbegin()) { // base case
			gcds[{rIds.rbegin()->p, cIds.rbegin()->p}] = K;
			continue;
		}
		// We have a rectangle which can be split in half in two ways
		// We can choose either arbitrarily, except for a 1xN or Nx1
		// rectangle, in which case there is only one choice.
		if (rit == rIds.rbegin()) {
			// cit->c1 or cit->c2 can be NO_ID if that child doesn't exist (hasn't been populated)
			// in the tree. This is not a problem, since in that case that key doesn't exist
			// in the map, so 0 is returned, as desired.
			gcds[{rit->p, cit->p}] =
				gcd2(getOrDefault(gcds, {rit->p, cit->c1}, 0LL), getOrDefault(gcds, {rit->p, cit->c2}, 0LL));
		} else {
			gcds[{rit->p, cit->p}] =
				gcd2(getOrDefault(gcds, {rit->c1, cit->p}, 0LL), getOrDefault(gcds, {rit->c2, cit->p}, 0LL));
		}
	}

}

long long calculate(int P, int Q, int U, int V) {
	vector<int> rIds = Rtree.decomposeIntervalAndGetIds({P, U+1});
	vector<int> cIds = Ctree.decomposeIntervalAndGetIds({Q, V+1});

	long long ans = 0;
	for(int rId : rIds)
	for(int cId : cIds) {
		ans = gcd2(ans, getOrDefault(gcds, {rId, cId}, 0LL));
		if (ans==1) { return 1; }
	}
	return ans;
}




inline long long gcd2(long long X, long long Y) {
	long long tmp;
	while (X != Y && Y != 0) {
		tmp = X;
		X = Y;
		Y = tmp % Y;
	}
	return X;
}

• Subtask 1: 10 / 10
• Subtask 2: 27 / 27
• Subtask 3: 26 / 26
• Subtask 4: 0 / 17
• Subtask 5: 0 / 20