Submission #612155

#	Time	Username	Problem	Language	Result	Execution time	Memory
612155		erekle	Connecting Supertrees (IOI20_supertrees)	C++17	100 / 100	296 ms	24132 KiB

supertrees

#include "supertrees.h"
#include <vector>

using namespace std;

int construct(vector<vector<int>> p) {
	int n = p.size();
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			if (p[i][j] == 3) return 0;
		}
	}

	// Split into components
	vector<vector<int>> components;
	vector<int> myCmp(n, -1);
	for (int i = 0; i < n; ++i) {
		if (myCmp[i] != -1) continue;
		components.push_back({});
		for (int j = 0; j < n; ++j) {
			if (p[i][j]) {
				if (myCmp[j] != -1) return 0;
				components.back().push_back(j);
				myCmp[j] = (int)components.size()-1;
			}
		}
	}

	// Check components are separate
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			if ((myCmp[i] == myCmp[j]) ^ (p[i][j] != 0)) return 0;
		}
	}

	// Build answer here
	vector<vector<int>> edge(n, vector<int>(n));

	// Now solve separately for each component
	for (vector<int> &c : components) {
		// Split into 1-chains
		vector<vector<int>> chains;
		vector<int> myChain(n, -1);
		for (int i : c) {
			if (myChain[i] != -1) continue;
			chains.push_back({});
			for (int j : c) {
				if (p[i][j] == 1) {
					if (myChain[j] != -1) return 0;
					chains.back().push_back(j);
					myChain[j] = (int)chains.size()-1;
				}
			}
		}

		// Check 1-chains are separate
		for (int i : c) {
			for (int j : c) {
				if ((myChain[i] == myChain[j]) ^ (p[i][j] == 1)) return 0;
			}
		}

		// Assign a chain leader to each chain
		int m = chains.size();
		vector<int> chainLeader(m);
		for (int i = 0; i < m; ++i) chainLeader[i] = chains[i][0];
		
		// Check chains and their leaders behave in the same way
		for (int i : c) {
			for (int j : c) {
				if (p[i][j] != p[chainLeader[myChain[i]]][j]) return 0;
			}
		}

		// There will be two ways to travel between a pair of nodes in different chains
		if (m == 2) return 0; // cannot have two chains (cannot make cycle)

		// Connect within chains
		for (int i : c) {
			if (chainLeader[myChain[i]] != i) {
				edge[i][chainLeader[myChain[i]]] = edge[chainLeader[myChain[i]]][i] = 1;
			}
		}

		// Connect chain leaders in a cycle
		if (m == 1) continue; // only one chain - no need
		for (int i = 0; i < m; ++i) {
			int j = (i+1)%m;
			edge[chainLeader[i]][chainLeader[j]] = edge[chainLeader[j]][chainLeader[i]] = 1;
		}
	}

	build(edge);
	return 1;
}

• Subtask 1: 11 / 11
• Subtask 2: 10 / 10
• Subtask 3: 19 / 19
• Subtask 4: 35 / 35
• Subtask 5: 21 / 21
• Subtask 6: 4 / 4