Submission #690529

#	Time	Username	Problem	Language	Result	Execution time	Memory
690529		Dan4Life	Road Closures (APIO21_roads)	C++17	12 / 100	47 ms	21580 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(),a.end()
using ll = long long;
const int maxn = (int)2e3+10;
const ll LINF = (ll)1e18;

int n;
ll dp[maxn][maxn];
vector<pair<int,ll>> adj[maxn];

void dfs(int s, int p){
    vector<pair<int,int>> v; v.clear();
    for(auto x : adj[s]){
        int u = x.fi, w = x.se;
        if(u==p) continue;
        dfs(u,s), v.pb({u,w});
        dp[s][0]+=dp[u][0]+w;
    }
    if(sz(v)==0) return;
    sort(all(v), [&](pair<int,int> a, pair<int,int> b){
         return a.se<b.se;});
    ll dp2[sz(v)+1][n+1];
    for(int i = 0; i <= sz(v); i++)
        for(int j = 0; j <= n; j++)
            dp2[i][j] = LINF;
    for(int i = 0; i < n; i++) dp2[0][i] = 0;
    for(int i = 1; i <= sz(v); i++){
        for(int j = 0; j < n; j++){
            int x = v[i-1].fi, y = v[i-1].se;
            if(j) dp2[i][j] = dp2[i][j-1];
            dp2[i][j]=min(dp2[i][j],dp2[i-1][j]+dp[x][j]+y);
            if(j) dp2[i][j]=min(dp2[i][j], dp2[i-1][j-1]+dp[x][j-1]);
        }
    }
    for(int k = 1; k < n; k++)
        dp[s][k] = min(dp[s][k-1], dp2[sz(v)][k]);
}

vector<ll> minimum_closure_costs(int N, vector<int> U, vector<int> V, vector<int> W) {
    n = N;
    vector<ll> ans(N,0ll);
    ans[0] = accumulate(all(W),0ll);
    if(accumulate(all(U),0ll)==0ll){
        sort(all(W)); reverse(all(W)); int i = 0;
        ll tot = accumulate(all(W),0ll);
        for(auto &u : ans) u=tot,tot-=W[i++];
        return ans;
    }
    bool isLine = 1;
    for(int i = 0; i < N-1; i++)
        isLine &= (U[i]==i and V[i]==i+1);
    if(isLine){
        ll dp[N-1][2];
        dp[0][0] = dp[0][1] = 0;
        for(int i = 1; i < N; i++){
            dp[i][0] = min(dp[i-1][1],dp[i-1][0])+W[i-1];
            dp[i][1] = dp[i-1][0];
        }
        ans[1] = min(dp[N-1][0],dp[N-1][1]);
        return ans;
    }
    for(int i = 0; i < N-1; i++){
        adj[U[i]].pb({V[i],W[i]});
        adj[V[i]].pb({U[i],W[i]});
    }
    dfs(0,-1); int i = 0;
    for(auto &u : ans) u = dp[0][i++];
    return ans;
}

• Subtask 1: 5 / 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