제출 #102233

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
102233		Benq	City (JOI17_city)	C++14	100 / 100	532 ms	53464 KiB

Encoder

#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>

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

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

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

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007; // 998244353
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

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); }

template<class A, class B> A operator+=(A& l, const B& r) { return l = l+r; }
template<class A, class B> A operator-=(A& l, const B& r) { return l = l-r; }
template<class A, class B> A operator*=(A& l, const B& r) { return l = l*r; }
template<class A, class B> A operator/=(A& l, const B& r) { return l = l/r; }

#include "Encoder.h"

static vi adj[1<<18];
static vi lens;

int dumb(int x) {
	return lb(all(lens),x)-lens.begin();
}

int dfs(int x, int y, int st) {
	int ST = st+1;
	trav(i,adj[x]) if (i != y) ST = dfs(i,x,ST);
	int d = dumb(ST-st); Code(x,((ll)st<<8)+d);
	return st+lens[d];
}

void Encode(int N, int A[], int B[]) {
	lens.pb(1);
	FOR(i,1,256) {
		int t = 1.05*lens.back();
		if (t == lens.back()) t ++;
		lens.pb(t);
	}
	F0R(i,N-1) adj[A[i]].pb(B[i]), adj[B[i]].pb(A[i]);
	dfs(0,-1,0);
}

Device

#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>

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

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

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

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define F0R(i, a) for (int i = 0; i < (a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= (a); i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define trav(a, x) for (auto& a : x)

#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound

#define sz(x) (int)x.size()
#define beg(x) x.begin()
#define en(x) x.end()
#define all(x) beg(x), en(x)
#define resz resize

const int MOD = 1000000007; // 998244353
const ll INF = 1e18;
const int MX = 100001;
const ld PI = 4*atan((ld)1);

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); }

template<class A, class B> A operator+=(A& l, const B& r) { return l = l+r; }
template<class A, class B> A operator-=(A& l, const B& r) { return l = l-r; }
template<class A, class B> A operator*=(A& l, const B& r) { return l = l*r; }
template<class A, class B> A operator/=(A& l, const B& r) { return l = l/r; }

#include "Device.h"

static vi lens;

void InitDevice() {
	lens.pb(1);
	FOR(i,1,256) {
		int t = 1.05*lens.back();
		if (t == lens.back()) t ++;
		lens.pb(t);
	}
}

pi decode(ll x) {
	pi a = {x>>8,lens[x%(1<<8)]};
	a.s += a.f;
	return a;
}

bool contain(pi a, pi b) {
	return a.f <= b.f && b.s <= a.s;
}

int Answer(long long S, long long T) {
	pi X = decode(S), Y = decode(T);
	if (contain(X,Y)) return 1;
	if (contain(Y,X)) return 0;
	return 2;
}

• Subtask 1: 8 / 8
• Subtask 2: 92 / 92