Submission #227077

#	Time	Username	Problem	Language	Result	Execution time	Memory
227077		rqi	Factories (JOI14_factories)	C++14	33 / 100	8039 ms	294152 KiB

factories

#include "factories.h"
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
typedef double db; 
typedef string str; 

typedef pair<int,int> pi;
typedef pair<ll,ll> pl; 
typedef pair<db,db> pd; 

typedef vector<int> vi; 
typedef vector<ll> vl; 
typedef vector<db> vd; 
typedef vector<str> vs; 
typedef vector<pi> vpi;
typedef vector<pl> vpl; 
typedef vector<pd> vpd; 

#define mp make_pair 
#define f first
#define s second
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) (x).rbegin(), (x).rend() 
#define rsz resize
#define ins insert 
#define ft front() 
#define bk back()
#define pf push_front 
#define pb push_back
#define eb emplace_back 
#define lb lower_bound 
#define ub upper_bound 

#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i,a) ROF(i,0,a)
#define trav(a,x) for (auto& a: x)

const int MOD = 1e9+7; // 998244353;
const int MX = 2e5+5; 
const ll INF = 1e18; 
const ld PI = acos((ld)-1);
const int xd[4] = {1,0,-1,0}, yd[4] = {0,1,0,-1}; 

template<class T> bool ckmin(T& a, const T& b) { 
	return b < a ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { 
	return a < b ? a = b, 1 : 0; } 
int pct(int x) { return __builtin_popcount(x); } 
int bit(int x) { return 31-__builtin_clz(x); } // floor(log2(x)) 
int cdiv(int a, int b) { return a/b+!(a<0||a%b == 0); } // division of a by b rounded up, assumes b > 0 

// INPUT
template<class A> void re(complex<A>& c);
template<class A, class B> void re(pair<A,B>& p);
template<class A> void re(vector<A>& v);
template<class A, size_t SZ> void re(array<A,SZ>& a);

template<class T> void re(T& x) { cin >> x; }
void re(db& d) { str t; re(t); d = stod(t); }
void re(ld& d) { str t; re(t); d = stold(t); }
template<class H, class... T> void re(H& h, T&... t) { re(h); re(t...); }

template<class A> void re(complex<A>& c) { A a,b; re(a,b); c = {a,b}; }
template<class A, class B> void re(pair<A,B>& p) { re(p.f,p.s); }
template<class A> void re(vector<A>& x) { trav(a,x) re(a); }
template<class A, size_t SZ> void re(array<A,SZ>& x) { trav(a,x) re(a); }

// TO_STRING
#define ts to_string
template<class A, class B> str ts(pair<A,B> p);
template<class A> str ts(complex<A> c) { return ts(mp(c.real(),c.imag())); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(char c) { str s = ""; s += c; return s; }
str ts(str s) { return s; }
str ts(const char* s) { return (str)s; }
str ts(vector<bool> v) { 
	bool fst = 1; str res = "{";
	F0R(i,sz(v)) {
		if (!fst) res += ", ";
		fst = 0; res += ts(v[i]);
	}
	res += "}"; return res;
}
template<size_t SZ> str ts(bitset<SZ> b) {
	str res = ""; F0R(i,SZ) res += char('0'+b[i]);
	return res; }
template<class T> str ts(T v) {
	bool fst = 1; str res = "{";
	for (const auto& x: v) {
		if (!fst) res += ", ";
		fst = 0; res += ts(x);
	}
	res += "}"; return res;
}
template<class A, class B> str ts(pair<A,B> p) {
	return "("+ts(p.f)+", "+ts(p.s)+")"; }

// OUTPUT
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { 
	pr(h); pr(t...); }
void ps() { pr("\n"); } // print w/ spaces
template<class H, class... T> void ps(const H& h, const T&... t) { 
	pr(h); if (sizeof...(t)) pr(" "); ps(t...); }

// DEBUG
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {
	cerr << to_string(h); if (sizeof...(t)) cerr << ", ";
	DBG(t...); }
#ifdef LOCAL // compile with -DLOCAL
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 42
#endif

// FILE I/O
void setIn(string s) { freopen(s.c_str(),"r",stdin); }
void setOut(string s) { freopen(s.c_str(),"w",stdout); }
void unsyncIO() { ios_base::sync_with_stdio(0); cin.tie(0); }
void setIO(string s = "") {
	unsyncIO();
	// cin.exceptions(cin.failbit); 
	// throws exception when do smth illegal
	// ex. try to read letter into int
	if (sz(s)) { setIn(s+".in"), setOut(s+".out"); } // for USACO
}

mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count()); 



/**
 * Description: 1D range minimum query. Can also do queries 
 	* for any associative operation in $O(1)$ with D\&C
 * Source: KACTL
 * Verification: 
	* https://cses.fi/problemset/stats/1647/
	* http://wcipeg.com/problem/ioi1223
	* https://pastebin.com/ChpniVZL
 * Memory: O(N\log N)
 * Time: O(1)
 */

template<class T> struct RMQ { // floor(log_2(x))
	int level(int x) { return 31-__builtin_clz(x); } 
	vector<T> v; vector<vi> jmp;
	int comb(int a, int b) { // index of min
		return v[a]==v[b]?min(a,b):(v[a]<v[b]?a:b); } 
	void init(const vector<T>& _v) {
		v = _v; jmp = {vi(sz(v))}; iota(all(jmp[0]),0);
		for (int j = 1; 1<<j <= sz(v); ++j) {
			jmp.pb(vi(sz(v)-(1<<j)+1));
			F0R(i,sz(jmp[j])) jmp[j][i] = comb(jmp[j-1][i],
									jmp[j-1][i+(1<<(j-1))]);
		}
	}
	int index(int l, int r) { // get index of min element
		int d = level(r-l+1);
		return comb(jmp[d][l],jmp[d][r-(1<<d)+1]); }
	T query(int l, int r) { return v[index(l,r)]; }
};


/**
 * Description: Euler Tour LCA w/ O(1) query. Compress takes a subset $S$ 
 	* of nodes and computes the minimal subtree that contains all the nodes 
	* pairwise LCA's and compressing edges. Returns a list of (par, orig\_index) 
	* representing a tree rooted at 0. The root points to itself.
 * Source: USACO, Simon Lindholm (KACTL)
 * Verification: USACO Debug the Bugs
 	* https://codeforces.com/contest/1320/problem/E
 */

template<int SZ> struct LCA {
	int N, R = 1, depth[SZ], st[SZ];
	ll dist[SZ];
	vector<pair<int, ll>> adj[SZ]; vpi tmp; RMQ<pi> r;
	void ae(int u, int v, ll d) { adj[u].pb(mp(v, d)), adj[v].pb(mp(u, d)); }
	void dfs(int u, int prev, ll d = 0){
		st[u] = sz(tmp), depth[u] = depth[prev]+1;
		dist[u] = d;
		tmp.pb({depth[u],u});
		trav(v,adj[u]) if (v.f != prev) 
			dfs(v.f,u,d+v.s), tmp.pb({depth[u],u});
	}
	void init(int _N) { N = _N; dfs(R,0); r.init(tmp); }
	int lca(int u, int v){
		u = st[u], v = st[v]; if (u > v) swap(u,v);
		return r.query(u,v).s; }
	ll getDist(int u, int v) {
		return dist[u]+dist[v]-2*dist[lca(u,v)]; }
	vpi compress(vi S) {
		static vi rev; rev.rsz(N+1);
		auto cmp = [&](int a, int b) { return st[a] < st[b]; };
		sort(all(S),cmp);
		int m = sz(S)-1; F0R(i,m) S.pb(lca(S[i],S[i+1]));
		sort(all(S),cmp); S.erase(unique(all(S)),end(S));
		F0R(i,sz(S)) rev[S[i]] = i;
		vpi ret = {pi(0,S[0])};
		F0R(i,sz(S)-1) ret.eb(rev[lca(S[i],S[i+1])],S[i+1]);
		return ret;
	}
};


/**
 * Description: The centroid of a tree of size $N$ is a vertex such that 
 	* after removing it, all resulting subtrees have size at most $\frac{N}{2}.$ 
 	* Can support tree path queries and updates.
 * Time: O(N\log N)
 * Source: own
 * Verification: 
	* https://codeforces.com/contest/342/problem/E
	* Triway Cup 2019 G
 */

template<int SZ> struct Centroid {
	vi adj[SZ];
	bool done[SZ]; // processed as centroid yet
	int sub[SZ], par[SZ]; // subtree size, current par
	pi cen[SZ]; // immediate centroid anc
	vi dist[SZ]; // dists to all centroid ancs
	void ae(int a, int b) { adj[a].pb(b), adj[b].pb(a); }
	void dfs(int x) {
		sub[x] = 1;
		trav(y,adj[x]) if (!done[y] && y != par[x]) {
			par[y] = x; dfs(y); sub[x] += sub[y]; }
	}
	int centroid(int x) {
		par[x] = -1; dfs(x); 
		for (int sz = sub[x];;) {
			pi mx = {0,0};
			trav(y,adj[x]) if (!done[y] && y != par[x]) 
				ckmax(mx,{sub[y],y});
			if (mx.f*2 <= sz) return x; 
			x = mx.s;
		}
	}
	void genDist(int x, int p) {
		dist[x].pb(dist[p].bk+1);
		trav(y,adj[x]) if (!done[y] && y != p) genDist(y,x); 
	} // CEN = {centroid above x, label of centroid subtree}
	void gen(pi CEN, int x) { 
		done[x = centroid(x)] = 1; cen[x] = CEN; 
		dist[x].pb(0); int co = 0;
		trav(y,adj[x]) if (!done[y]) genDist(y,x);
		trav(y,adj[x]) if (!done[y]) gen({x,co++},y);
	}
	void init() { gen({-1,0},1); } // start with vertex 1
};



const int mx = 500005;

LCA<mx> tr;
Centroid<mx> cen;
int N;
ll shortest[mx][2];

void Init(int _N, int A[], int B[], int D[]) {
	N = _N;
	for(int i = 0; i <= N-2; i++){
		int a = A[i];
		int b = B[i];
		a++;
		b++;
		ll d = D[i];
		tr.ae(a, b, d);
		cen.ae(a, b);
	}
	tr.init(N+2);
	cen.init();
	for(int i = 1; i <= N; i++){
		shortest[i][0] = shortest[i][1] = INF;
	}
}

ll ans;

long long Query(int S, int X[], int T, int Y[]) {

	for(int i = 0; i < S; i++) X[i]++;
	for(int i = 0; i < T; i++) Y[i]++;
	vi nodes;
	//ps(shortest[4][1]);
	for(int i = 0; i < S; i++){
		int node = X[i];
		while(node != -1){

			ckmin(shortest[node][0], tr.getDist(X[i], node));
			nodes.pb(node);
			node = cen.cen[node].f;
		}
	}
	for(int i = 0; i < T; i++){
		int node = Y[i];
		while(node != -1){
			ckmin(shortest[node][1], tr.getDist(Y[i], node));
			nodes.pb(node);
			node = cen.cen[node].f;
		}
	}

	ans = INF;
	for(auto u: nodes){
		ckmin(ans, shortest[u][0]+shortest[u][1]);
	}

	/*for(auto u: nodes){
		ps(u, shortest[u][0], shortest[u][1]);
	}*/
	//reset arrays based on nodes
	for(auto u: nodes){
		shortest[u][0] = shortest[u][1] = INF;
	}
	return ans;
}


/*int main() {
	setIO();
	
	// you should actually read the stuff at the bottom
}*/

/* stuff you should look for
	* int overflow, array bounds
	* special cases (n=1?)
	* do smth instead of nothing and stay organized
	* WRITE STUFF DOWN
*/

Compilation message (stderr)

factories.cpp: In function 'void setIn(std::__cxx11::string)':
factories.cpp:124:31: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
 void setIn(string s) { freopen(s.c_str(),"r",stdin); }
                        ~~~~~~~^~~~~~~~~~~~~~~~~~~~~
factories.cpp: In function 'void setOut(std::__cxx11::string)':
factories.cpp:125:32: warning: ignoring return value of 'FILE* freopen(const char*, const char*, FILE*)', declared with attribute warn_unused_result [-Wunused-result]
 void setOut(string s) { freopen(s.c_str(),"w",stdout); }
                         ~~~~~~~^~~~~~~~~~~~~~~~~~~~~~

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