제출 #1056340

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1056340		Muhammet	도로 폐쇄 (APIO21_roads)	C++17	5 / 100	34 ms	9676 KiB

roads

#include <bits/stdc++.h>
#include "roads.h"

using namespace std;

#define ll long long
#define ff first
#define ss second

vector<ll> minimum_closure_costs(int n, vector<int> u, vector<int> u1, vector<int> w) {
    vector <pair<ll,pair<ll,ll>>> v;
    ll s = 0;
    bool tr = 0;
    for(int i = 0; i < n-1; i++){
        v.push_back({w[i],{u[i],u1[i]}});
        if(u1[i] != u[i]+1) tr = 1;
        s += w[i];
    }
    vector <ll> v1;
    sort(v.rbegin(), v.rend());
    if(*max_element(u.begin(), u.end()) == 0){
        ll s2 = s;
        v1.push_back(s);
        for(int i = 0; i < n-1; i++){
            s2 -= (v[i].ff);
            v1.push_back(s2);
        }
        return v1;
    }
    if(*max_element(w.begin(), w.end()) == 1){
        vector <ll> p(n,0), v2, p1(n,0), v1;
        for(int i = 0; i < n-1; i++){
            p[u[i]]++;
            p[u1[i]]++;
        }
        for(int i = 0; i < n; i++){
            v2.push_back(p[i]);
        }
        sort(v2.begin(), v2.end());
        p1[0] = v2[0];
        for(int i = 1; i < n; i++){
            p1[i] = p1[i-1] + v2[i];
        }
        int ind = 0;
        for(int i = 0; i < n; i++){
            while(v2[ind] <= i and ind < n){
                ind++;
            }
            ll x = 0;
            if(ind != 0) x = p[ind-1];
            v1.push_back(p[n-1]-(x+(n-ind)*i));
        }
        return v1;
    }
    if(tr == 0){
        v1.push_back(s);
        vector <pair<int,int>> v2;
        for(int i = 0; i < n-1; i++){
            v2.push_back({u[i],w[i]});
        }
        sort(v2.begin(), v2.end());
        vector <ll> p1(n+1,0), p2(n+1,0);
        for(int i = 0; i < n-1; i++){
            if(i > 0) p1[i] = p1[i-1], p2[i] = p2[i-1];
            if(i % 2 == 0) p2[i] += v2[i].ss;
            else p1[i] += v2[i].ss;
        }
        ll mx = max(p1[n-2],p2[n-2]);
        for(int i = 0; i < n-2; i++){
            mx = max(mx,p1[i]+p2[n-2]-p2[min(i+2,n-2)]);
            mx = max(mx,p2[i]+p1[n-2]-p1[min(i+2,n-2)]);
        }
        v1.push_back(s-mx);
        for(int i = 2; i < n; i++) v1.push_back(0);
        return v1;
    }
    return vector<ll>(n-1,0);
}

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