Submission #1200592

#	Time	Username	Problem	Language	Result	Execution time	Memory
1200592		countless	Road Closures (APIO21_roads)	C++20	7 / 100	521 ms	9620 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;

const ll MOD = 998244353;
const ll INF = 1e18;
const ld EPS = 1e-12;

#define endl "\n"
#define sp <<" "<<
#define REP(i, a, b) for(ll i = a; i < b; i++)
#define dbg(x) cout << #x << " = " << x << endl
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fast_io() ios_base::sync_with_stdio(false); cin.tie(NULL)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define sz(x) ((ll)(x).size())

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        // http://xorshift.di.unimi.it/splitmix64.c
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }

    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

template <typename Key, typename Value>
using hash_map = unordered_map<Key, Value, custom_hash>;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// uniform_int_distribution<int>(a, b)(rng);
// shuffle(all(a), rng);

vector<ll> minimum_closure_costs(int N, vector<int> U, vector<int> V, vector<int> W) {
    vector<ll> ans(N, 0);

    // k == 0
    ll sum = 0;
    REP(i, 0, N-1) {
        sum += W[i];
    }
    ans[0] = sum;

    // k == 1
    vector<pair<ll, ll>> dp(N, {INF, INF});
    dp[1] = {W[0], 0};
    REP(i, 2, N) {
        // take
        dp[i].first = min(dp[i-1].first, dp[i-1].second) + W[i-1];
        
        // no take
        dp[i].second = dp[i-1].first;

        cerr << i sp W[i-1] sp dp[i].first sp dp[i].second << endl;
    }

    ans[1] = min(dp[N-1].first, dp[N-1].second);

    return ans;
}

• Subtask 1: 0 / 5
• Subtask 2: 7 / 7
• Subtask 3: 0 / 14
• Subtask 4: 0 / 10
• Subtask 5: 0 / 17
• Subtask 6: 0 / 25
• Subtask 7: 0 / 22