Codeforces Round #595 (Div. 3) Editorial

All ideas belong to MikeMirzayanov.

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		cin >> n;
		vector<int> a(n);
		for (int j = 0; j < n; ++j) {
			cin >> a[j];
		}
		bool ok = true;
		for (int p1 = 0; p1 < n; ++p1) {
			for (int p2 = 0; p2 < p1; ++p2) {
				ok &= abs(a[p1] - a[p2]) != 1;
			}
		}
		cout << 2 - ok << endl;
	}
	
	return 0;
}

1249B1 - Обмен книгами (простая версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		cin >> n;
		vector<int> p(n);
		for (int j = 0; j < n; ++j) {
			cin >> p[j];
			--p[j];
		}
		for (int j = 0; j < n; ++j) {
			int cnt = 0;
			int k = j;
			do {
				++cnt;
				k = p[k];
			} while (k != j);
			cout << cnt << " ";
		}
		cout << endl;
	}
	
	return 0;
}

1249B2 - Обмен книгами (сложная версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		cin >> n;
		vector<int> p(n);
		for (int j = 0; j < n; ++j) {
			cin >> p[j];
			--p[j];
		}
		vector<int> used(n);
		vector<int> ans(n);
		for (int j = 0; j < n; ++j) {
			if (!used[j]) {
				vector<int> cur;
				while (!used[j]) {
					cur.push_back(j);
					used[j] = true;
					j = p[j];
				}
				for (auto el : cur) ans[el] = cur.size();
			}
		}
		for (int j = 0; j < n; ++j) cout << ans[j] << " ";
		cout << endl;
	}
	
	return 0;
}

1249C1 - Хорошие числа (простая версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		int n;
		cin >> n;
		while (true) {
			bool ok = true;
			int cur = n;
			while (cur > 0) {
				if (ok && cur % 3 == 2) ok = false;
				cur /= 3;
			}
			if (ok) break;
			++n;
		}
		cout << n << endl;
	}
	
	return 0;
}

1249C2 - Хорошие числа (сложная версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const long long INF64 = 1e18;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int q;
	cin >> q;
	for (int i = 0; i < q; ++i) {
		long long n;
		cin >> n;
		vector<int> vals;
		int pos2 = -1;
		while (n > 0) {
			vals.push_back(n % 3);
			if (vals.back() == 2) pos2 = int(vals.size()) - 1;
			n /= 3;
		}
		vals.push_back(0);
		if (pos2 != -1) {
			int pos0 = find(vals.begin() + pos2, vals.end(), 0) - vals.begin();
			vals[pos0] = 1;
			for (int i = pos0 - 1; i >= 0; --i) {
				vals[i] = 0;
			}
		}
		long long ans = 0;
		long long pw = 1;
		for (int i = 0; i < int(vals.size()); ++i) {
			ans += pw * vals[i];
			pw *= 3;
		}
		if (vals.back() == 1) ans = pw / 3;
		cout << ans << endl;
	}
	
	return 0;
}

1249D1 - Слишком много отрезков (простая версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int N = 250;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, k;
	cin >> n >> k;
	vector<pair<int, int>> segs(n);
	vector<int> cnt(N);
	for (int i = 0; i < n; ++i) {
		cin >> segs[i].first >> segs[i].second;
		++cnt[segs[i].first];
		--cnt[segs[i].second + 1];
	}
	
	for (int i = 0; i + 1 < N; ++i) {
		cnt[i + 1] += cnt[i];
	}
	
	vector<int> ans(n);
	for (int i = 0; i < N; ++i) {
		while (cnt[i] > k) {
			int pos = -1;
			for (int p = 0; p < n; ++p) {
				if (!ans[p] && (segs[p].first <= i && i <= segs[p].second) && (pos == -1 || segs[p].second > segs[pos].second)) {
					pos = p;
				}
			}
			assert(pos != -1);
			for (int j = segs[pos].first; j <= segs[pos].second; ++j) {
				--cnt[j];
			}
			ans[pos] = 1;
		}
	}
	
	cout << accumulate(ans.begin(), ans.end(), 0) << endl;
	for (int i = 0; i < n; ++i) {
		if (ans[i]) cout << i + 1 << " ";
	}
	cout << endl;
	
	return 0;
}

1249D2 - Слишком много отрезков (сложная версия)

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int N = 200200;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, k;
	cin >> n >> k;
	vector<pair<int, int>> segs(n);
	vector<int> cnt(N);
	vector<vector<int>> ev(N);
	for (int i = 0; i < n; ++i) {
		cin >> segs[i].first >> segs[i].second;
		++cnt[segs[i].first];
		--cnt[segs[i].second + 1];
		ev[segs[i].first].push_back(i + 1);
		ev[segs[i].second + 1].push_back(-i - 1);
	}
	
	for (int i = 0; i + 1 < N; ++i) {
		cnt[i + 1] += cnt[i];
	}
	
	vector<int> ans(n), sub(N);
	set<pair<int, int>> curSegs;
	int curSub = 0;																																				
	for (int i = 0; i < N; ++i) {
		curSub += sub[i];
		for (auto it : ev[i]) {
			if (it > 0) {
				curSegs.insert(make_pair(segs[it - 1].second, it - 1));
			} else {
				auto iter = curSegs.find(make_pair(segs[-it - 1].second, -it - 1));
				if (iter != curSegs.end()) curSegs.erase(iter);
			}
		}
		while (cnt[i] - curSub > k) {
			assert(!curSegs.empty());
			int pos = curSegs.rbegin()->second;
			curSegs.erase(prev(curSegs.end()));
			++curSub;
			--sub[segs[pos].second + 1];
			ans[pos] = 1;
		}
	}
	
	cout << accumulate(ans.begin(), ans.end(), 0) << endl;
	for (int i = 0; i < n; ++i) {
		if (ans[i]) cout << i + 1 << " ";
	}
	cout << endl;
	
	return 0;
}

1249E - На лифте или по лестнице?

Tutorial

Solution

#include <bits/stdc++.h>

using namespace std;

const int INF = 1e9;

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	int n, c;
	cin >> n >> c;
	vector<int> a(n - 1), b(n - 1);
	for (int i = 0; i < n - 1; ++i) {
		cin >> a[i];
	}
	for (int i = 0; i < n - 1; ++i) {
		cin >> b[i];
	}
	vector<vector<int>> dp(n, vector<int>(2, INF));
	dp[0][0] = 0, dp[0][1] = c;
	for (int i = 0; i < n - 1; ++i) {
		dp[i + 1][0] = min(dp[i + 1][0], dp[i][1] + a[i]);
		dp[i + 1][0] = min(dp[i + 1][0], dp[i][0] + a[i]);
		dp[i + 1][1] = min(dp[i + 1][1], dp[i][1] + b[i]);
		dp[i + 1][1] = min(dp[i + 1][1], dp[i][0] + b[i] + c);
	}
	
	for (int i = 0; i < n; ++i) {
		cout << min(dp[i][0], dp[i][1]) << " ";
	}
	cout << endl;
	
	return 0;
}

1249F - Подмножество максимального веса

Tutorial

Solution (Vovuh, n^3)

#include <bits/stdc++.h>

using namespace std;

const int N = 210;

int n, k;
vector<int> a;
vector<vector<int>> g, dp;

void dfs(int v, int p) {
	dp[v][0] = a[v];
	for (auto to : g[v]) {
		if (to != p) dfs(to, v);
	}
	for (int dep = 0; dep < N; ++dep) {
		if (dep == 0) {
			for (auto to : g[v]) {
				if (to == p) continue;
				dp[v][dep] += dp[to][max(0, k - dep - 1)];
			}
		} else {
			for (auto to : g[v]) {
				if (to == p) continue;
				int cur = dp[to][dep - 1];
				for (auto other : g[v]) {
					if (other == p || other == to) continue;
					cur += dp[other][max(dep - 1, k - dep - 1)];
				}
				dp[v][dep] = max(dp[v][dep], cur);
			}
		}
	}
	for (int dep = N - 1; dep > 0; --dep) {
		dp[v][dep - 1] = max(dp[v][dep - 1], dp[v][dep]);
	}
}

int main() {
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);
#endif
	
	cin >> n >> k;
	++k;
	a = vector<int>(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
	}
	g = vector<vector<int>>(n);
	for (int i = 0; i < n - 1; ++i) {
		int x, y;
		cin >> x >> y;
		--x, --y;
		g[x].push_back(y);
		g[y].push_back(x);
	}
	
	dp = vector<vector<int>>(n, vector<int>(N));
	dfs(0, -1);
	cout << dp[0][0] << endl;
	
	return 0;
}

Solution (PikMike, n^2)

#include <bits/stdc++.h>

using namespace std;

#define forn(i, n) for (int i = 0; i < int(n); ++i)

const int N = 200 + 13;
const int INF = 1e9;

int n, k;
int a[N];
vector<int> g[N];
int dp[N][N];
int d[N];
int tmp[N];

void dfs(int v, int p = -1){
	d[v] = 1;
	dp[v][0] = a[v];
	for (auto u : g[v]) if (u != p){
		dfs(u, v);
		int nw = max(d[v], d[u] + 1);
		forn(i, nw)
			tmp[i] = -INF;
		forn(i, d[v]) for (int j = max(0, k - i); j < d[u]; ++j)
			tmp[min(i, j + 1)] = max(tmp[min(i, j + 1)], dp[v][i] + dp[u][j]);
		forn(i, d[u])
			tmp[i + 1] = max(tmp[i + 1], dp[u][i]);
		forn(i, d[v])
			dp[v][i] = max(dp[v][i], tmp[i]);
		for (int i = d[v]; i < nw; ++i)
			dp[v][i] = tmp[i];
		d[v] = nw;
		for (int i = d[v] - 1; i > 0; --i)
			dp[v][i - 1] = max(dp[v][i - 1], dp[v][i]);
	}
}

int main(){
	scanf("%d%d", &n, &k);
	forn(i, n)
		scanf("%d", &a[i]);
	forn(i, n - 1){
		int v, u;
		scanf("%d%d", &v, &u);
		--v, --u;
		g[v].push_back(u);
		g[u].push_back(v);
	}
	dfs(0);
	printf("%d\n", dp[0][0]);
}

Rev.	By	When	Δ	Comment
en3	vovuh	2019-10-24 12:56:09	159
ru2	vovuh	2019-10-24 12:55:11	148
en2	vovuh	2019-10-22 22:43:39	2	(published)
ru1	vovuh	2019-10-22 22:43:33	10164	Первая редакция перевода на Русский
en1	vovuh	2019-10-22 22:42:18	10194	Initial revision (saved to drafts)

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