제출 #736363

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
736363		hoainiem	도로 폐쇄 (APIO21_roads)	C++14	36 / 100	551 ms	66140 KiB

roads

#include "roads.h"
#include <bits/stdc++.h>
#define fi first
#define se second
#define nmax 100008
const long long inf = 1e18;
using namespace std;
typedef pair<long long, long long> pii;
long long n, sub1 = true, sub2 = true;
long long dp[2008][2008][2];
long long f[5][nmax][2];
vector<pii> l[nmax];
long long sdq(int id, int x, int pre, int kt){
    if (f[id][x][kt] != -1)
        return f[id][x][kt];
    if (!id && kt)
        return f[id][x][kt] = inf;
    long long res = 0;
    vector<long long>vt;
    for (pii it : l[x])
        if (it.fi != pre){
            res += sdq(id, it.fi, x, 1);
            vt.push_back(sdq(id, it.fi, x, 0) + it.se - sdq(id, it.fi, x, 1));
        }
    sort(vt.begin(), vt.end(), greater<long long>());
    while (!vt.empty() && ((int)vt.size() + kt > id || vt.back() < 0)){
        res += vt.back();
        vt.pop_back();
    }
    return f[id][x][kt] = res;
}
long long dq(int id, int x, int pre, int kt){
    if (dp[id][x][kt] != -1)
        return dp[id][x][kt];
    if (!id && kt)
        return dp[id][x][kt] = inf;
    long long res = 0;
    vector<long long>vt;
    for (pii it : l[x])
        if (it.fi != pre){
            res += dq(id, it.fi, x, 1);
            vt.push_back(dq(id, it.fi, x, 0) + it.se - dq(id, it.fi, x, 1));
        }
    sort(vt.begin(), vt.end(), greater<long long>());
    while (!vt.empty() && ((int)vt.size() + kt > id || vt.back() < 0)){
        res += vt.back();
        vt.pop_back();
    }
    return dp[id][x][kt] = res;
}
std::vector<long long> minimum_closure_costs(int N, std::vector<int> U, std::vector<int> V, std::vector<int> W) {
    n = N;
    for (long long i = 0; i < n - 1; i++){
        if (U[i] != 0)
            sub1 = false;
        if (U[i] != i || V[i] != i + 1)
            sub2 = false;
        l[U[i]].push_back({V[i], W[i]});
        l[V[i]].push_back({U[i], W[i]});
    }
    vector<long long>res;
    if (sub1){
        sort(W.begin(), W.end(), greater<int>());
        long long j = 0;
        for (int tmp : W)
            j += tmp;
        res.push_back(j);
        for (long long i = 0; i < n - 1; i++){
            j -= W[i];
            res.push_back(j);
        }
        return res;
    }
    if (sub2){
        memset(f, -1, sizeof(f));
        for (long long k = 0; k < n; k++)
            if (k < 2)
                res.push_back(sdq(k, 0, 0, 0));
            else
                res.push_back(0);
        return res;
    }
    assert(n <= 2000);
    memset(dp, -1, sizeof(dp));
    for (long long k = 0; k < n; k++)
        res.push_back(dq(k, 0, 0, 0));
    return res;
}

• Subtask 1: 5 / 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