제출 #558157

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
558157		Neos	공장들 (JOI14_factories)	C++17	15 / 100	8058 ms	146796 KiB

factories

#include"factories.h"
#include <bits/stdc++.h>
using namespace std;

using ll = long long;
using ld = long double;
using ull = unsigned long long;
using ii = pair<int, int>;

#define fi first
#define se second
#define pb push_back
#define numBit(x) (__builtin_popcountll(1ll * (x)))
#define getBit(x, i) ((x) >> (i) & 1)
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()

template<class X, class Y>
	bool minimize(X &x, const Y &y) {
		X eps = 1e-9;
		if (x > y + eps) {
			x = y;
			return true;
		} else return false;
	}

template<class X, class Y>
	bool maximize(X &x, const Y &y) {
		X eps = 1e-9;
		if (x + eps < y) {
			x = y;
			return true;
		} else return false;
	}

const int N = 5e5 + 7, LOG = 20;
long long oo = 1e18 + 7;
//sample program

int n, m, sz[N], par[N], P[N][LOG];
ll ans[N], d[N], h[N];
bool done[N];
vector<ii> adj[N];

void get_sz(int u, int par) {
	sz[u] = 1;
	for (auto [v, w]: adj[u]) if (v != par && !done[v]) {
		get_sz(v, u); sz[u] += sz[v];
	}
}

int centroid(int u, int par, int n) {
	for (auto [v, w]: adj[u]) if (v != par && !done[v] && sz[v] * 2 > n)
		return centroid(v, u, n);
	return u;
}

int solve(int u) {
	get_sz(u, -1); int rt = centroid(u, -1, sz[u]); done[rt] = 1;
	for (auto [v, w]: adj[rt]) if (!done[v])
        par[solve(v)] = rt;
	return rt;
}

void dfs(int u) {
    for (int i = 1; i < LOG; i++) P[u][i] = P[P[u][i - 1]][i - 1];
    for (auto [v, w]: adj[u]) if (v != P[u][0])
        P[v][0] = u, h[v] = h[u] + 1, d[v] = d[u] + w, dfs(v);
}

int lca(int u, int v) {
    if (h[u] < h[v]) swap(u, v);
    for (int i = LOG - 1; i >= 0; i--)
        if (h[u] - (1 << i) >= h[v]) u = P[u][i];
    if (u == v) return u;
    for (int i = LOG - 1; i >= 0; i--)
        if (P[u][i] && P[u][i] != P[v][i])
            u = P[u][i], v = P[v][i];
    return P[u][0];
}

long long dis(int u, int v) {
    return d[u] + d[v] - 2 * d[lca(u, v)];
}

void Init(int N, int A[], int B[], int D[]){
	n = N; fill(ans, ans + n, oo);
	for (int i = 0; i < n - 1; i++) {
		adj[A[i]].pb(ii(B[i], D[i])); adj[B[i]].pb(ii(A[i], D[i]));
	}

    par[solve(0)] = -1; dfs(0);
}

void upd(int u, bool t) {
	for (int v = u; v >= 0; v = par[v])
		if (t) minimize(ans[v], dis(u, v)); else ans[v] = oo;
}

long long get(int u) {
	long long res = oo;
	for (int v = u; v >= 0; v = par[v]) minimize(res, ans[v] + dis(u, v));
	return res;
}

long long Query(int S, int X[], int T, int Y[]) {
	long long res = 1e18;
	for (int i = 0; i < S; i++) upd(X[i], 1);
	for (int i = 0; i < T; i++) minimize(res, get(Y[i]));
	for (int i = 0; i < S; i++) upd(X[i], 0);
	return res;
}

• Subtask 1: 15 / 15
• Subtask 2: 0 / 18
• Subtask 3: 0 / 67