제출 #1056896

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1056896		c2zi6	기지국 (IOI20_stations)	C++14	21 / 100	502 ms	948 KiB

stations

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define ff first
#define ss second
#define pb push_back
#define all(a) (a).begin(), (a).end()
#define replr(i, a, b) for (int i = int(a); i <= int(b); ++i)
#define reprl(i, a, b) for (int i = int(a); i >= int(b); --i)
#define rep(i, n) for (int i = 0; i < int(n); ++i)
#define mkp(a, b) make_pair(a, b)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VPI;
typedef vector<VI> VVI;
typedef vector<VVI> VVVI;
typedef vector<VPI> VVPI;
typedef pair<ll, ll> PLL;
typedef vector<ll> VL;
typedef vector<PLL> VPL;
typedef vector<VL> VVL;
typedef vector<VVL> VVVL;
typedef vector<VPL> VVPL;
template<class T> T setmax(T& a, T b) {if (a < b) return a = b; return a;}
template<class T> T setmin(T& a, T b) {if (a < b) return a; return a = b;}
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;
template<class T>
using indset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#include "stations.h"

VVI gp;
VI ret;
int last;

void dfs(int u, int p = -1) {
    ret[u] = last++;
    for (int v : gp[u]) if (v != p) {
        dfs(v, u);
        if (p == -1) last += 1000;
    }
}

VI label(int n, int k, VI U, VI V) {
    gp = VVI(n);
    rep(i, n-1) {
        int u = U[i];
        int v = V[i];
        gp[u].pb(v);
        gp[v].pb(u);
    }
    ret = VI(n);
    last = 0;
    int s = -1;
    rep(u, n) if (gp[u].size() > 2) s = u;
    else if (gp[u].size() == 1 && s == -1) s = u;
    dfs(s);
    
    /*for (int x : ret) cout << x << " "; cout << endl;*/
    return ret;
}

int find_next_station(int s, int t, VI c) {
    if (c.size() == 1) return c[0];
    if (s == 0) {
        rep(i, c.size()) {
            if (c[i]%1000 > t%1000) return c[i-1];
        }
        return c.back();
    }
    if (t/1000 != s/1000) return c[0];
    if (t > s) return c[1];
    return c[0];
}

• Subtask 1: 5 / 5
• Subtask 2: 0 / 8
• Subtask 3: 16 / 16
• Subtask 4: 0 / 10
• Subtask 5: 0 / 61