oj.uz

Submission #410889

# Submission time Handle Problem Language Result Execution time Memory
410889 2021-05-23T21:38:25 Z 534351 Stations (IOI20_stations) C++17
0 / 100
 1077 ms 904 KB 
#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, compress;
//inside each vtx, we store start time and finish time.

void dfs(int u, int p)
{
	st[u] = T;
	T++;
	for (int v : edge[u])
	{
		if (v == p) continue;
        depth[v] = depth[u] + 1;
		dfs(v, u);
	}
    ft[u] = T;
    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.clear();
	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);
	ans.resize(N);
    FOR(i, 0, N)
    {
        if (depth[i] % 2)
        {
            ans[i] = ft[i];
        }
        else
        {
            ans[i] = st[i];
        }
    }
    compress = ans;
    sort(ALL(compress));
    FOR(i, 0, N)
    {
        ans[i] = LB(ALL(compress), ans[i]) - compress.begin();
    }
	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)
        {
            st = ch[0] + 1;
        }
        else
        {
            st = ch[0] - 1;
        }
        if (st <= v && v <= u)
        {
            //you're going to an st, which is <= v.
            FORD(i, SZ(ch), 0)
            {
                if (ch[i] <= v) return ch[i];
            }
        }
        else
        {
            return ch[0];
        }
    }
    assert(false);
}

Subtask #1
0 / 5.0

# Verdict Execution time Memory Grader output
1 Runtime error 5 ms 792 KB Execution killed with signal 6
2 Halted 0 ms 0 KB -

Subtask #2
0 / 8.0

# Verdict Execution time Memory Grader output
1 Runtime error 61 ms 776 KB Execution killed with signal 6
2 Halted 0 ms 0 KB -

Subtask #3
0 / 16.0

# Verdict Execution time Memory Grader output
1 Runtime error 10 ms 900 KB Execution killed with signal 6
2 Halted 0 ms 0 KB -

Subtask #4
0 / 10.0

# Verdict Execution time Memory Grader output
1 Correct 1077 ms 504 KB Output is correct
2 Correct 840 ms 400 KB Output is correct
3 Correct 671 ms 528 KB Output is correct
4 Correct 3 ms 608 KB Output is correct
5 Runtime error 2 ms 860 KB Execution killed with signal 6
6 Halted 0 ms 0 KB -

Subtask #5
0 / 61.0

# Verdict Execution time Memory Grader output
1 Runtime error 15 ms 904 KB Execution killed with signal 6
2 Halted 0 ms 0 KB -