제출 #99001

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
99001		qkxwsm	공장들 (JOI14_factories)	C++14	33 / 100	8058 ms	208376 KiB

factories

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/rope>
#include "factories.h"

using namespace std;
using namespace __gnu_pbds;
using namespace __gnu_cxx;

random_device(rd);
mt19937 rng(rd());
const long long FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();

struct custom_hash
{
	template<class T>
	unsigned long long operator()(T v) const
	{
		unsigned long long x = v;
		x += FIXED_RANDOM;
		// x += 11400714819323198485ull;
		// x = (x ^ (x >> 30)) * 13787848793156543929ull;
		x = (x ^ (x >> 27)) * 10723151780598845931ull;
		return x ^ (x >> 31);
	}
};

template<class T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<class T, class U> using hash_table = gp_hash_table<T, U, custom_hash>;

template<class T>
T randomize(T mod)
{
	return (uniform_int_distribution<T>(0, mod - 1))(rng);
}
template<class T>
void readi(T &x)
{
	x = 0;
	bool negative = false;
	char c = ' ';
	while (c < '-')
	{
		c = getchar();
	}
	if (c == '-')
	{
		negative = true;
		c = getchar();
	}
	while (c >= '0')
	{
		x = x * 10 + (c - '0');
		c = getchar();
	}
	if (negative)
	{
		x = -x;
	}
}
template<class T>
void printi(T output)
{
	if (output == 0)
	{
		putchar('0');
		return;
	}
	if (output < 0)
	{
		putchar('-');
		output = -output;
	}
	int buf[20], n = 0;
	while(output)
	{
		buf[n] = ((output % 10));
		output /= 10;
		n++;
	}
	for (n--; n >= 0; n--)
	{
		putchar(buf[n] + '0');
	}
	return;
}
template<class T>
void ckmin(T &a, T b)
{
	a = min(a, b);
}
template<class T>
void ckmax(T &a, T b)
{
	a = max(a, b);
}
long long expo(long long a, long long e, long long mod)
{
	return ((e == 0) ? 1 : ((expo(a * a % mod, e >> 1, mod)) * ((e & 1) ? a : 1) % mod));
}
template<class T, class U>
void nmod(T &x, U mod)
{
	if (x >= mod) x -= mod;
}
template<class T>
T gcd(T a, T b)
{
	return (b ? gcd(b, a % b) : a);
}

#define y0 ___y0
#define y1 ___y1
#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#define DBG(x) cerr << #x << " = " << (x) << endl
#define SZ(x) ((int) ((x).size()))
#define FOR(i, a, b) for (auto i = (a); i < (b); i++)
#define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--)
#define ALL(x) (x).begin(), (x).end()

const long double PI = 4.0 * atan(1.0);
const long double EPS = 1e-9;

#define MAGIC 347
#define SINF 10007
#define CO 1000007
#define INF 1000000007
#define BIG 1000000931
#define LARGE 1696969696967ll
#define GIANT 2564008813937411ll
#define LLINF 2696969696969696969ll
#define MAXN 500013

typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ld, ld> pdd;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<ld> vd;
typedef vector<pii> vpi;
typedef vector<pll> vpl;
typedef vector<pdd> vpd;

int N, M, T;
vpl edge[MAXN], edge1[MAXN];
int parent[MAXN], depth[MAXN];
int parent1[MAXN];
int mintable[25][2 * MAXN];
int ord[2 * MAXN], rord[2 * MAXN];
int st[MAXN], ft[MAXN];
ll dis[MAXN];
ll dp[2][MAXN];
vi L, R, V;
vpi pts;
ll ans;

int comb1(int a, int b)
{
	return (depth[ord[a]] < depth[ord[b]] ? a : b);
}
int lca(int u, int v)
{
	u = rord[u]; v = rord[v];
	if (u > v) swap(u, v);
	int sz = 31 - __builtin_clz(v - u + 1);
	int res = ord[comb1(mintable[sz][u], mintable[sz][v - (1 << sz) + 1])];
	return res;
}
void dfs(int u)
{
	ord[M] = u;
	rord[u] = M;
	M++;
	st[u] = T;
	ft[u] = T;
	T++;
	for (pii p : edge[u])
	{
		int v = p.se;
		if (v == parent[u]) continue;
		parent[v] = u;
		depth[v] = depth[u] + 1;
		dis[v] = dis[u] + p.fi;
		dfs(v);
		ft[u] = ft[v];
		ord[M] = u;
		M++;
	}
}
ll dist(int u, int v)
{
	// cerr << u << ' ' << v << ' ' << lca(u, v) << endl;
	return dis[u] + dis[v] - 2 * dis[lca(u, v)];
}
void Init(int n, int A[], int B[], int D[])
{
	N = n;
	FOR(i, 0, N - 1)
	{
		int u = A[i], v = B[i], d = D[i];
		edge[u].PB({d, v});
		edge[v].PB({d, u});
	}
	parent[0] = N;
	dfs(0);
	FOR(i, 0, M)
	{
		mintable[0][i] = i;
	}
	FOR(j, 1, 21)
	{
		FOR(i, 0, M)
		{
			mintable[j][i] = mintable[j - 1][i];
			if (i + (1 << (j - 1)) < M)
			{
				mintable[j][i] = comb1(mintable[j][i], mintable[j - 1][i + (1 << (j - 1))]);
			}
		}
	}
	FOR(i, 0, N)
	{
		dp[0][i] = dp[1][i] = LLINF;
		parent1[i] = N;
	}
}
bool cmp(int a, int b)
{
	return st[a] < st[b];
}
bool cmp1(pii a, pii b)
{
	return st[a.fi] < st[b.fi];
}
int ok(int idx)
{
	int u = V[idx];
	int e;
	for (e = idx + 1; e < SZ(V) && st[u] <= st[V[e]] && st[V[e]] <= ft[u]; )
	{
		int v = V[e];
		ll d = dis[v] - dis[u];
		edge1[u].PB({d, v});
		edge1[v].PB({d, u});
		parent1[v] = u;
		e = ok(e);
	}
	return e;
}


long long Query(int S, int X[], int T, int Y[])
{
	FOR(i, 0, S)
	{
		L.PB(X[i]);
	}
	FOR(i, 0, T)
	{
		R.PB(Y[i]);
	}
	for (int u : L)
	{
		pts.PB({u, 0});
	}
	for (int u : R)
	{
		pts.PB({u, 1});
	}
	sort(ALL(pts), cmp1);
	FOR(i, 0, SZ(pts))
	{
		V.PB(pts[i].fi);
		if (i)
		{
			int x = lca(pts[i - 1].fi, pts[i].fi);
			if (x != pts[i].fi && x != pts[i - 1].fi)
			{
				V.PB(x);
			}
		}
	}
	sort(ALL(V), cmp);
	// V.erase(unique(ALL(V)), V.end());
	ok(0);
	priority_queue<pll, vector<pll>, greater<pll> > pq;
	for (int u : L)
	{
		dp[0][u] = 0;
		pq.push({0, u});
	}
	while(!pq.empty())
	{
		ll d = pq.top().fi, u = pq.top().se;
		pq.pop();
		if (d != dp[0][u]) continue;
		for (pll p : edge1[u])
		{
			int u = p.se; ll nd = d + p.fi;
			if (nd < dp[0][u])
			{
				dp[0][u] = nd;
				pq.push({nd, u});
			}
		}
	}
	for (int u : R)
	{
		dp[1][u] = 0;
		pq.push({0, u});
	}
	while(!pq.empty())
	{
		ll d = pq.top().fi, u = pq.top().se;
		pq.pop();
		if (d != dp[1][u]) continue;
		for (pll p : edge1[u])
		{
			int u = p.se; ll nd = d + p.fi;
			if (nd < dp[1][u])
			{
				dp[1][u] = nd;
				pq.push({nd, u});
			}
		}
	}
	ans = LLINF;
	for (int u : V)
	{
		ckmin(ans, dp[0][u] + dp[1][u]);
	}
	for (int u : V)
	{
		dp[0][u] = LLINF;
		dp[1][u] = LLINF;
	}
	for (int u : V)
	{
		edge1[u].clear();
		parent1[u] = N;
	}
	L.clear();
	R.clear();
	pts.clear();
	V.clear();
	//u need to build the tree!
	return ans;
}

/* READ READ READ
* int overflow, maxn too small, special cases (n=1?, two distinct?), cin.tie() interactive
* reread the problem, try small cases
* note down possible sources of error as you go
* do smth instead of nothing
*/

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