Submission #361984

#	Time	Username	Problem	Language	Result	Execution time	Memory
361984		ryansee	Stations (IOI20_stations)	C++14	100 / 100	1124 ms	1312 KiB

stations

#include "stations.h"
 
#include "bits/stdc++.h"
using namespace std;
 
#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define f first
#define s second
#define cbr cerr<<"hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x,y) (lower_bound(all(x),y)-x.begin())
#define ubd(x,y) (upper_bound(all(x),y)-x.begin())
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
inline long long rand(long long x, long long y) { return rng() % (y+1-x) + x; } //inclusive
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }
 
using ll=long long; 
using ld=long double;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
using pi=pair<ll,ll>; using spi=pair<ll,pi>; using dpi=pair<pi,pi>; 
 
long long LLINF = 1e18;
int INF = 1e9+1e6;
#define MAXN (200006)
 
vector<int> label(int n, int k, vector<int> U, vector<int> V) {
	vector<vector<int>> v(n, vector<int>());
	FOR(i,0,n-2) v[U[i]].eb(V[i]), v[V[i]].eb(U[i]);
	vector<int> st(n, 0), en(n, 0);
	int co = 0;
	function<void(int,int)>dfs=[&](int x,int p){
		st[x]=co++;
		for(auto i:v[x]) if(i^p) dfs(i, x);
		en[x]=co++;
	};
	dfs(0, 0);
	vector<int> C(n);
	function<void(int,int,int)>dfs2=[&](int x,int p,int co){
		if(co == 0) C[x] = st[x];
		else C[x] = en[x];
		for(auto i:v[x]) if(i^p) dfs2(i, x, !co);
	};
	dfs2(0, 0, 0);
	vector<int> d;
	for(auto i:C) d.eb(i);
	sort(all(d));
	for(auto&i:C) i=lbd(d, i);
	return C;
}
int find_next_station(int S, int T, vector<int> c) {
	if(siz(c) == 1) return c[0];
	if(S == 0) {
		for(auto i:c) if(i >= T) return i;
		assert(0);
	}
	int pre, post, b;
	if(c[0] < S) { // S is post order
		post = S, pre = c[1] - 1, b = 1;
	} else { // S is pre
		pre = S, post = c[siz(c)-2] + 1, b = 0;
	}
	if(pre <= T && T <= post) { // in subtree
		if(b==0) {
			for(auto i:c) if(i >= T) return i;
			assert(0);
		} else {
			reverse(all(c));
			for(auto i:c) if(i <= T) return i;
			assert(0);
		}
	} else {
		if(b) return c[0];
		else return c.back();
	}
}

• Subtask 1: 5 / 5
• Subtask 2: 8 / 8
• Subtask 3: 16 / 16
• Subtask 4: 10 / 10
• Subtask 5: 61 / 61