Submission #979170

#	Time	Username	Problem	Language	Result	Execution time	Memory
979170		Mher777	Road Closures (APIO21_roads)	C++17	7 / 100	63 ms	22264 KiB

roads

#include "roads.h"
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <iomanip>
#include <array>
#include <string>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <bitset>
#include <list>
#include <iterator>
#include <numeric>
#include <complex>
#include <utility>
#include <random>
#include <cassert>
#include <fstream>
using namespace std;
mt19937 rnd(7069);
typedef int itn;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef float fl;
typedef long double ld;
using vi = vector<int>;
using vll = vector<ll>;
using mii = map<int, int>;
using mll = map<ll, ll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
#define ff first
#define ss second
#define pub push_back
#define pob pop_back
#define puf push_front
#define pof pop_front
#define mpr make_pair
#define yes cout<<"Yes\n"
#define no cout<<"No\n"
#define all(x) (x).begin(), (x).end()
const int dx[8] = { -1, 0, 1, 0, -1, -1, 1, 1 };
const int dy[8] = { 0, -1, 0, 1, -1, 1, -1, 1 };
const int MAX = int(2e9 + 5);
const ll MAXL = ll(1e18) + 5ll;
const ll MOD = ll(1000000007);
const ll MOD2 = ll(998244353);

const int N = 100005;
ll dp[N];
vector<pair<int, ll>> g[N];
vll v;
int n;

void dfs(int u, int par) {
	for (auto to : g[u]) {
		if (to.ff == par) continue;
		v.pub(to.ss);
		dfs(to.ff, u);
	}
}

vll minimum_closure_costs(int NX, vi U, vi V, vi W) {
	n = NX;
	vll ans(n);
	int u = -1;
	for (int i = 0; i < n - 1; ++i) {
		++U[i];
		++V[i];
		ans[0] += W[i];
		g[U[i]].pub({ V[i],W[i] });
		g[V[i]].pub({ U[i],W[i] });
	}
	for (int i = 1; i <= n; ++i) {
		if ((int)g[i].size() == 1) u = i;
	}
	dfs(u, u);
	int m = (int)v.size();
	dp[0] = v[0];
	for (int i = 1; i < m; ++i) {
		if (i == 1) {
			dp[i] = v[i];
			continue;
		}
		dp[i] = v[i] + min(dp[i - 1], dp[i - 2]);
	}
	if (m > 1) {
		ans[1] = min(dp[m - 1], dp[m - 2]);
	}
	return ans;
}

/*
2
0 1 1
*/

• 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