Submission #1106588

#	Time	Username	Problem	Language	Result	Execution time	Memory
1106588		Octagons	Sphinx's Riddle (IOI24_sphinx)	C++17	6.50 / 100	788 ms	2200 KiB

sphinx

#include <bits/stdc++.h>
using namespace std;
#include "sphinx.h"
#define sz(x) int(x.size())
int n, mmo;
int p[255], sz[255], p2[255], sz2[255], dep[255], idxo = 1;
vector<int> adj57[255], tr[255], bck[255], adj2[255], nds[255];
int vis[255];
bool fnd[255];
 
int findp2(int u) {
    return p2[u] = (p2[u] == u ? u : findp2(p2[u]));
}
 
bool merge2(int u, int v) {
    u = findp2(u), v = findp2(v);
    if (u == v)return false;
    if (sz2[u] < sz2[v])swap(u, v);
    sz2[u] += sz2[v];
    p2[v] = u;
	return true;
}
 
int findp(int u) {
    return p[u] = (p[u] == u ? u : findp(p[u]));
}
 
void merge(int u, int v) {
    u = findp(u), v = findp(v);
    if (u == v)return;
    if (sz[u] < sz[v])swap(u, v);
    sz[u] += sz[v];
    p[v] = u;
}
 
bool chk(vector<int> v) {
	for (int i = 0; i < n; i++) {
		p2[i] = i;
		sz2[i] = 1;
	}
	int comps = n;
	for (int i = 0; i < n; i++) {
		for (auto &vv : adj57[i]) {
			if (~v[i] && v[i] == v[vv]) {
				comps -= merge2(i, vv);
			} else if (v[i] == -1 && v[vv] == -1 && findp(i) == findp(vv)) {
				comps -= merge2(i, vv);
			}
		}
	}
	int x = perform_experiment(v);
	return (comps > x);
}
 
bool chk2(int u, int v) {
	for (int i = 0; i < n; i++) {
		p2[i] = i;
		sz2[i] = 1;
	}
	int comps = n;
	vector<int> e(n, n);
	e[u] = e[v] = -1;
	for (int i = 0; i < n; i++) {
		for (auto &vv : adj57[i]) {
			if (e[i] == e[vv]) {
				comps -= merge2(i, vv);
			}
		}
	}
	int x = perform_experiment(e);
	return (comps == x);
}
 
void init(int u, int pr) {
	vis[u] = idxo++;
	for (auto &v : adj57[u]) {
		if (v == pr)continue;
		if (vis[v]) {
			if (vis[v] < vis[u])bck[u].push_back(v);
			continue;
		}
		init(v, u);
		tr[u].push_back(v);
	}
}
 
void soly(int u, vector<int> vvo) {
	if (vvo.empty())return;
	int l = 0, r = int(vvo.size()) - 1, id = -1;
	while (l <= r) {
		int m = (l + r) >> 1;
		vector<int> e(n, n);
		for (int j = 0; j <= m; j++) {
			for (int i = 0; i < n; i++) {
				if (findp(i) == vvo[j])e[i] = -1;
			}
		}
		e[u] = -1;
		if (chk(e)) {
			id = m;
			r = m-1;
		} else {
			l = m+1;
		}
	}
	if (~id) {
		merge(u, vvo[id]);
	} else {
		return;
	}
	vector<int> xx;
	for (int i = id+1; i < sz(vvo); i++) {
		xx.push_back(vvo[i]);
	}
	soly(u, xx);
}
 
void dfs(int u) {
	vector<int> vvo;
	for (auto &v : bck[u]) {
		if (findp(v) != findp(u))vvo.push_back(findp(v));
	}
	sort(vvo.begin(), vvo.end());
	vvo.erase(unique(vvo.begin(), vvo.end()), vvo.end());
	soly(u, vvo);
	for (auto &v : tr[u]) {
		if (chk2(u, v)) {
			merge(u, v);
		}
		dfs(v);
	}
}

void init2(int u, int pr) {
	vis[u] = 1;
	for (auto &v : adj2[u]) {
		if (vis[v])continue;
		dep[v] = (dep[u] ^ 1);
		init2(v, u);
	}
}

vector<int> nw(vector<int> x, int u) {
	vector<int> ret;
	for (auto &i : x) {
		if (i == u)continue;
		ret.push_back(i);
	}
	return ret;
}

std::vector<int> find_colours(int NH, std::vector<int> X, std::vector<int> Y) {
    n = NH;
    vector<int> ret(n, 0);
    for (int i = 0; i < n; i++) {
      p[i] = i;
      sz[i] = 1;
	  adj2[i].clear();
	  adj57[i].clear();
	  fnd[i] = false;
	  nds[i].clear();
	  tr[i].clear();
	  bck[i].clear();
	  vis[i] = 0;
    }
    for (int i = 0; i < X.size(); i++) {
      int u = X[i], v = Y[i];
      adj57[u].push_back(v);
      adj57[v].push_back(u);
    }
	mmo = X.size();
	init(0, 0);
	dfs(0);
	bool ff = true;
    for (int i = 0; i < n; i++) {
      ret[i] = findp(i);
	  ff &= (findp(i) == findp(0));
    }
	if (ff) {
		for (int i = 0; i < n; i++) {
			vector<int> e(n, n);
			e[0] = -1;
			e[adj57[0][0]] = i;
			if (chk(e)) {
				for (int j = 0; j < n; j++) {
					ret[j] = i;
				}
				break;
			}
		}
		return ret;
	}
	for (int i = 0; i < n; i++) {
		nds[findp(i)].push_back(i);
		for (auto &v : adj57[i]) {
			if (findp(i) == findp(v))continue;
			adj2[findp(i)].push_back(findp(v));
			adj2[findp(v)].push_back(findp(i));
		}
		vis[i] = 0;
		dep[i] = 0;
	}
	init2(findp(0), -1);
	vector<int> fo, so;
	for (int i = 0; i < n; i++) {
		if (findp(i) != i)continue;
		if (dep[i])fo.push_back(i);
		else so.push_back(i);
	}
	for (int i = 0; i < n; i++) {
		vector<int> cnd;
		for (auto &j : fo) {
			if (fnd[j])continue;
			cnd.push_back(j);
		}
		while (!cnd.empty()) {
			int id = -1, l = 0, r = sz(cnd) - 1;
			while (l <= r) {
				int m = (l + r) >> 1;
				vector<int> e(n, n);
				for (auto &j : so) {
					e[j] = i;
				}
				for (int j = 0; j <= m; j++) {
					for (auto &jj : nds[cnd[j]]) {
						e[jj] = -1;
					}
				}
				if (chk(e)) {
					id = m;
					r = m-1;
				} else {
					l = m+1;
				}
			}
			if (id == -1)break;
			fnd[cnd[id]] = true;
			for (int j = 0; j < n; j++) {
				if (findp(j) == cnd[id]) {
					ret[j] = i;
				}
			}
			vector<int> xx;
			for (int j = id+1; j < sz(cnd); j++) {
				xx.push_back(cnd[j]);
			}
			cnd = xx;
		}
	}
	swap(fo, so);
	for (int i = 0; i < n; i++) {
		dep[i] ^= 1;
	}
	for (int i = 0; i < n; i++) {
		vector<int> cnd;
		for (auto &j : fo) {
			if (fnd[j])continue;
			cnd.push_back(j);
		}
		while (!cnd.empty()) {
			int id = -1, l = 0, r = sz(cnd) - 1;
			while (l <= r) {
				int m = (l + r) >> 1;
				vector<int> e(n, n);
				for (auto &j : so) {
					e[j] = i;
				}
				for (int j = 0; j <= m; j++) {
					for (auto &jj : nds[cnd[j]]) {
						e[jj] = -1;
					}
				}
				if (chk(e)) {
					id = m;
					r = m-1;
				} else {
					l = m+1;
				}
			}
			if (id == -1)break;
			fnd[cnd[id]] = true;
			for (int j = 0; j < n; j++) {
				if (findp(j) == cnd[id]) {
					ret[j] = i;
				}
			}
			vector<int> xx;
			for (int j = id+1; j < sz(cnd); j++) {
				xx.push_back(cnd[j]);
			}
			cnd = xx;
		}
	}
    return ret;
}

Compilation message (stderr)

sphinx.cpp: In function 'std::vector<int> find_colours(int, std::vector<int>, std::vector<int>)':
sphinx.cpp:166:23: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<int>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
  166 |     for (int i = 0; i < X.size(); i++) {
      |                     ~~^~~~~~~~~~

• Subtask 0: 0 / 0
• Subtask 1: 3 / 3
• Subtask 2: 3.5 / 7
• Subtask 3: 0 / 33
• Subtask 4: 0 / 21
• Subtask 5: 0 / 36