Submission #1126709

#	Time	Username	Problem	Language	Result	Execution time	Memory
1126709		czauderna	Road Closures (APIO21_roads)	C++20	31 / 100	2095 ms	31816 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define st first
#define nd second
#define pb push_back
const int NN = 1e5+3;
const ll INF = 1e16;
vector<int> inp[NN], Vdeg[NN], grf[NN];
int pre[NN], cnt;
bool odw[NN];
ll Dp[NN][2], C[NN];

// dla n^2 log
vector<ll> tmpX, tmpY;
bool comp(int a, int b){
    return (int)inp[a].size() > (int)inp[b].size();
}
void set_pre(int v, int p){
    pre[v]=cnt; cnt++;
    for(auto w:inp[v]){
        if(w==p) continue;
        set_pre(w, v);
    }
}
int get_x(int v, int k){
    int out=0;
    for(auto w:inp[v]) if((int)inp[w].size()<=k && pre[v]<pre[w]) out++, tmpX.pb(C[w]);
    sort(tmpX.begin(), tmpX.end());
    return out;
}
ll get_y(int v, int k){
    ll out = 0;
    for(auto w:inp[v]) if((int)inp[w].size()>k && pre[v]<pre[w]) out+=Dp[w][0], tmpY.pb(Dp[w][1]-Dp[w][0]);
    sort(tmpY.begin(), tmpY.end());
    return out;
}
ll get_sm_x(int x){
    ll out=0;
    for(int i=0; i<x; i++) out += tmpX[i];
    return out;
}
ll get_sm_y(int x){
    ll out=0;
    for(int i=0; i<x; i++) out += tmpY[i];
    return out;
}
void get_out(int v, int k){
    odw[v]=1;
    for(auto w:grf[v]){
        if(odw[w]) continue;
        get_out(w, k);
    }
    tmpX.clear(); tmpY.clear();
    int diff = (int)inp[v].size()-k;
    int X = get_x(v, k); 
    ll smY = get_y(v, k);
    int Y = (int)tmpY.size();
    for(int i=0; i<=Y; i++){
        if(X+i<diff-1) continue;
        Dp[v][1]=min(Dp[v][1], get_sm_x(max(diff-i-1, 0))+smY+get_sm_y(i)+C[v]);
        if(X+i<diff) continue;
        Dp[v][0]=min(Dp[v][0], get_sm_x(max(diff-i, 0))+smY+get_sm_y(i));
        Dp[v][1]=min(Dp[v][1], get_sm_x(max(diff-i, 0))+smY+get_sm_y(max(i-1, 0))+C[v]);
    }
}
vector<ll> minimum_closure_costs(int N, vector<int> U, vector<int> V, vector<int> W) {
    //PREPROCES
    int n=N;
    for(int i=0; i<n-1; i++){
        inp[U[i]].pb(V[i]);
        inp[V[i]].pb(U[i]);
    }
    for(int i=0; i<n; i++) Dp[i][0]=Dp[i][1]=INF;
    for(int i=0; i<n; i++) Vdeg[(int)inp[i].size()].pb(i);
    set_pre(0, 0);
    for(int i=0; i<n-1; i++){
        if(pre[U[i]]<pre[V[i]]) C[V[i]] = W[i];
        else C[U[i]] = W[i];
    }
    C[0] = INF;
    for(int i=0; i<n; i++) sort(inp[i].begin(), inp[i].end(), comp);

    //WYNIK
    vector<pair<int, int>> Vert;
    vector<ll> WYNIK;
    for(int k=n-1; k>=0; k--){
        ll out=0;
        for(auto v:Vdeg[k+1]){
            Vert.pb({pre[v], v});
            for(auto w:inp[v]){
                if(inp[w].size()>=k+1) grf[v].pb(w);
                if(inp[w].size()>k+1) grf[w].pb(v);
            }
        }
        sort(Vert.begin(), Vert.end());
        for(auto Ve:Vert){
            int v=Ve.nd;
            if(odw[v]) continue;
            get_out(v, k);
            out+=min(Dp[v][0], Dp[v][1]);
        }
        WYNIK.pb(out);
        //CZYSZCZENIE
        for(auto v:Vert){
            odw[v.nd]=0;
            Dp[v.nd][0]=Dp[v.nd][1]=INF;
        }
    }
    reverse(WYNIK.begin(), WYNIK.end());
    return WYNIK;
}

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