제출 #410902

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
410902		534351	기지국 (IOI20_stations)	C++17	100 / 100	1160 ms	816 KiB

stations

#include "stations.h"
#include <bits/stdc++.h>

using namespace std;

template<class T, class U>
void ckmin(T &a, U b)
{
    if (a > b) a = b;
}

template<class T, class U>
void ckmax(T &a, U b)
{
    if (a < b) a = b;
}

#define MP make_pair
#define PB push_back
#define LB lower_bound
#define UB upper_bound
#define fi first
#define se second
#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 SZ(x) ((int) (x).size())
#define ALL(x) (x).begin(), (x).end()

const int MAXN = 1013;
const int INF = 1e9 + 7;

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

int N, K, T;
vi edge[MAXN];
int st[MAXN], ft[MAXN], depth[MAXN];
vi ans;
//inside each vtx, we store start time and finish time.

void dfs(int u, int p)
{
    if (depth[u] % 2 == 0)
    {
        st[u] = T;
        ans[u] = st[u];
        T++;
    }
	for (int v : edge[u])
	{
		if (v == p) continue;
        depth[v] = depth[u] + 1;
		dfs(v, u);
	}
    if (depth[u] % 2 == 1)
    {
        ft[u] = T;
        ans[u] = ft[u];
        T++;
    }
}

vi label(int n, int k, vi U, vi V)
{
	N = n; K = k; T = 0;
	FOR(i, 0, N)
	{
		edge[i].clear();
	}
    ans.resize(N);
	FOR(i, 0, N - 1)
	{
		int u = U[i], v = V[i];
		edge[u].PB(v);
		edge[v].PB(u);
	}
    depth[0] = 0;
	dfs(0, N);
	return ans;
}

int find_next_station(int u, int v, vi ch)
{
    sort(ALL(ch));
    //figure out if you're a st or ft.
    //case 1: you're an st. then any neighbor will have ft > u.
    if (u < ch[0])
    {
        int ft = -1;
        //you're an st. your ft is the second largest neighbor unless you are the root, and then it's the largest + 1.
        if (u == 0)
        {
            ft = ch.back() + 1;
        }
        else
        {
            ft = (SZ(ch) > 1 ? ch[SZ(ch) - 2] : u) + 1;
        }
        if (u <= v && v <= ft)
        {
            //you're going to an ft, which still is >= v.
            FOR(i, 0, SZ(ch))
            {
                if (ch[i] >= v) return ch[i];
            }
        }
        else
        {
            return ch.back();
        }
    }
    else
    {
        //you're an ft. your st is the smallest child - 1, unless you only have a parent.
        int st = -1;
        if (SZ(ch) == 1) //you are a leaf. then whatever.
        {
            return ch[0];
        }
        else //you have some children. the minimum guy to be considered in your subtree must be ch[0]
        {
            st = ch[0];
        }
        if (st <= v && v <= u)
        {
            //you're going to an st in your subtree
            FORD(i, SZ(ch), 0)
            {
                if (ch[i] <= v) return ch[i];
            }
        }
        else
        {
            return ch[0];
        }
    }
    assert(false);
}

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