Submission #1124383

#	Time	Username	Problem	Language	Result	Execution time	Memory
1124383		ArtistWall	Toll (APIO13_toll)	C++20	100 / 100	2118 ms	5820 KiB

toll

#include <bits/stdc++.h>
#define long long long

using namespace std;

struct union_find {
	vector<int> s, p;

	union_find(int n = 0) : s(n, 1), p(n) {
		iota(p.begin(), p.end(), 0);
	}

	int find(int u) {
		if (p[u] != u) {
			p[u] = find(p[u]);
		}
		return p[u];
	}

	bool merge(int u, int v) {
		u = find(u);
		v = find(v);
		if (u == v) {
			return false;
		}
		if (s[u] < s[v]) {
			swap(u, v);
		}
		p[v] = u;
		s[u] = s[u] + s[v];
		return true;
	}

	bool same(int u, int v) {
		return find(u) == find(v);
	}

	int size(int u) {
		u = find(u);
		return s[u];
	}
};

const int N = (int) 1E5;
const int M = (int) 3E5;
const int K = 20;
const int A = numeric_limits<int>::max();

int n, m, k;

bool t1[M];
array<int, 3> e1[M];
array<int, 2> e2[K];
array<int, 3> e3[K];
vector<int> e4[K + 1];

int a[N];
long e[K + 1];

void Kruskal1() {
	union_find s(n);
	for (int i = 0; i < k; i++) {
		int u = e2[i][0], v = e2[i][1];
		s.merge(u, v);
	}
	for (int i = 0; i < m; i++) {
		int u = e1[i][1], v = e1[i][2];
		t1[i] = s.merge(u, v);
	}
}

int root;

void Kruskal2() {
	union_find s1(n);
	for (int i = 0; i < m; i++) {
		int u = e1[i][1], v = e1[i][2];
		if (t1[i] == true) {
			s1.merge(u, v);
		}
	}
	int c = 0;
	vector<int> p(n, -1);
	for (int i = 0; i < n; i++) {
		if (s1.find(i) == i) {
			p[i] = c;
			c = c + 1;
		}
	}
	for (int i = 0; i < k; i++) {
		int u = e2[i][0], v = e2[i][1];
		u = p[s1.find(u)];
		v = p[s1.find(v)];
		e2[i] = {u, v};
	}
	int t = 0;
	union_find s2(k + 1);
	for (int i = 0; i < m; i++) {
		int u = e1[i][1], v = e1[i][2], w = e1[i][0];
		if (t1[i] == false) {
			int x = p[s1.find(u)];
			int y = p[s1.find(v)];
			if (s2.merge(x, y)) {
				e3[t] = {w, x, y};
				t = t + 1;
			}
		}
	}
	assert(t + 1 == c);
	for (int i = 0; i < n; i++) {
		int u = p[s1.find(i)];
		e[u] = e[u] + a[i];
	}
	root = p[s1.find(0)];
}

int p[K + 1], d[K + 1];
int g[K + 1];
long t[K + 1];

void DFS(int u) {
	t[u] = e[u];
	for (auto v : e4[u]) {
		if (v != p[u]) {
			p[v] = u;
			d[v] = d[u] + 1;
			DFS(v);
			t[u] = t[u] + t[v];
		}
	}
}

void check(int u, int v, int w) {
	while (u != v) {
		if (d[u] > d[v]) {
			swap(u, v);
		}
		g[v] = min(g[v], w);
		v = p[v];
	}
}

void solve() {
	cin >> n >> m >> k;
	for (int i = 0; i < m; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		u = u - 1, v = v - 1;
		e1[i] = {w, u, v};
	}
	for (int i = 0; i < k; i++) {
		int u, v;
		cin >> u >> v;
		u = u - 1, v = v - 1;
		e2[i] = {u, v};
	}
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}

	sort(e1, e1 + m);
	Kruskal1();
	Kruskal2();


	long c = 0;
	for (int S = 0; S < (1 << k); S++) {
		union_find s(k + 1);
		bool f = true;
		for (int i = 0; i < k; i++) {
			if (S >> i & 1) {
				int u = e2[i][0], v = e2[i][1];
				if (not s.merge(u, v)) {
					f = false;
					break;
				}
			}
		}
		if (not f) {
			continue;
		}

		vector<bool> t3(k);
		for (int i = 0; i < k; i++) {
			int u = e3[i][1], v = e3[i][2];
			t3[i] = s.merge(u, v);
		}

		for (int i = 0; i < k + 1; i++) {
			e4[i] = vector<int>();
		}
		for (int i = 0; i < k; i++) {
			if (S >> i & 1) {
				int u = e2[i][0], v = e2[i][1];
				e4[u].push_back(v);
				e4[v].push_back(u);
			}
		}
		for (int i = 0; i < k; i++) {
			if (t3[i] == true) {
				int u = e3[i][1], v = e3[i][2];
				e4[u].push_back(v);
				e4[v].push_back(u);
			}
		}

		p[root] = -1;
		d[root] = 0;
		DFS(root);
		fill(g, g + k + 1, A);
		for (int i = 0; i < k; i++) {
			if (t3[i] == false) {
				int u = e3[i][1], v = e3[i][2], w = e3[i][0];
				check(u, v, w);
			}
		}

		long b = 0;
		for (int i = 0; i < k; i++) {
			if (S >> i & 1) {
				int u = e2[i][0], v = e2[i][1];
				int x = d[u] > d[v] ? u : v;
				b = b + t[x] * g[x];
			}
		}

		c = max(c, b);
	}
	cout << c << "\n";
}

int32_t main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int t;
	t = 1;
	for (int i = 0; i < t; i++) {
		solve();
	}
	return 0;
}

• Subtask 1: 16 / 16
• Subtask 2: 18 / 18
• Subtask 3: 22 / 22
• Subtask 4: 22 / 22
• Subtask 5: 22 / 22