AtCoder Regular Contest 146 D >=<

洛谷传送门

AtCoder 传送门

考虑直接增量构造。

把条件拆成形如 \(a_u \ge x\) 时 \(a_v \gets \max(a_v, y)\)，连边，跑类似一个 spfa 的东西，就行了。

这样一定能构造出一组和最小的解。

启示：二元关系，一般考虑建图。

code

// Problem: D - >=<
// Contest: AtCoder - AtCoder Regular Contest 146
// URL: https://atcoder.jp/contests/arc146/tasks/arc146_d
// Memory Limit: 1024 MB
// Time Limit: 4000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 1000100;

ll n, m, q, a[maxn], head[maxn], len;
bool vis[maxn];

struct E {
	ll v, x, y;
	E(ll a = 0, ll b = 0, ll c = 0) : v(a), x(b), y(c) {}
};

vector<E> G[maxn];

void solve() {
	scanf("%lld%lld%lld", &n, &m, &q);
	while (q--) {
		ll u, v, x, y;
		scanf("%lld%lld%lld%lld", &u, &x, &v, &y);
		G[u].pb(v, x, y);
		G[v].pb(u, y, x);
		G[u].pb(v, x + 1, y + 1);
		G[v].pb(u, y + 1, x + 1);
	}
	queue<int> q;
	for (int i = 1; i <= n; ++i) {
		a[i] = 1;
		q.push(i);
		vis[i] = 1;
		sort(G[i].begin(), G[i].end(), [&](E a, E b) {
			return a.x > b.x;
		});
	}
	while (q.size()) {
		int u = q.front();
		q.pop();
		vis[u] = 0;
		while (G[u].size() && a[u] >= G[u].back().x) {
			E e = G[u].back();
			G[u].pop_back();
			ll v = e.v, y = e.y;
			if (a[v] < y) {
				a[v] = y;
				if (!vis[v]) {
					vis[v] = 1;
					q.push(v);
				}
			}
		}
	}
	ll ans = 0;
	for (int i = 1; i <= n; ++i) {
		if (a[i] > m) {
			puts("-1");
			return;
		}
		ans += a[i];
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}