제출 #976662

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
976662		josanneo22	도로 폐쇄 (APIO21_roads)	C++17	24 / 100	2075 ms	23992 KiB

roads

#include "roads.h"

#include <vector>

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;
#define vi vector<int>

vector<i64> minimum_closure_costs(int N, vi U, vi V, vi W) {
    vector<vector<pair<int, int>>> G(N);
    for (int i = 0; i < N - 1; i++) {
        G[U[i]].push_back(make_pair(V[i], W[i]));
        G[V[i]].push_back(make_pair(U[i], W[i]));
    }
    auto solve = [&](int k) {
        // cout << k << '\n';
        vector<vector<i64>> dp(N, vector<i64>(2, 100000));
        function<void(int, int)> dfs = [&](int u, int f) {
            // cout << "tree : " << u << ' ' << f << '\n';
            // if (G[u].size() < k) {
            //     dp[u][0] = dp[u][1] = 0;
            // }
            i64 sum = 0;
            vector<i64> wait; 
            for (auto & v : G[u]) {
                if (v.first == f) continue;
                dfs(v.first, u);
                wait.push_back(-dp[v.first][1] + dp[v.first][0] + v.second);
                sum += dp[v.first][0] + v.second;
                // cout << v.first << ' ' << dp[v.first][0] + v.second << ' ' << dp[v.first][1] << '\n';
            }
            // cout << "total: " << sum << '\n';
            sort(wait.begin(), wait.end(), greater<i64>());
            // cout << "wait : ";
            // for (auto & v : wait) cout << v << ' ';
            //     cout << '\n';
            int deg = G[u].size();
            int remain = deg - max(deg - k - 1, 0);
            // cout << "remain: " << remain << '\n';
            i64 res = sum, S = sum;
            for (int i = 0; i < min(remain, (int)wait.size()); i++) {
                S -= wait[i];
                res = min(res, S);
            }
            dp[u][0] = res;
            remain = deg - max(deg - k, 0);
            res = sum; S = sum;
            // cout << "remain: " << remain << '\n';
            // cout << "sums : ";
            for (int i = 0; i < min(remain, (int)wait.size()); i++) {
                S -= wait[i];
                res = min(res, S);
            }
            // cout << '\n';
            dp[u][1] = res;
            // cout << "++++++++++++++\n";
        };  
        dfs(0, -1);
        // for (int i = 0; i < N; i++) {
        //     cout << "dp : " << i << ' ' << dp[i][0] << ' ' << dp[i][1] << '\n';
        // }
        return dp[0][0];
    };  
    vector<i64> ans(N);
    for (int k = 0; k < N; k++) { 
        ans[k] = solve(k - 1);
    }
    return ans;
}

/*
对于每个K, 我们可以单独solve
就是说如果我们让dp[i] = 整个i的子树里最小价格满足, deg[i] <= k
发现到其实这样的存储方式不足以满足dp转移方式
dp[u][0/1] = 和父亲节点是否有边
    如果我们拿这个边那么我们会得到dp[v][1]
    如果我们不要这个边我们得到dp[v][0] + W[i]
        如果我们现在和父亲节点连边: 需要移除min(deg[u] - k, 0)
                   没和父亲连边： 需要移除min(deg[u] - k - 1, 0)
                   那么我们可以拿的边 = deg[u] - 移除
    我们先把dp[v][1] - (dp[v][0] + W[i])排序， ans = sum(dp[v][0] + W[i])
    然后我们降序排序然后直接贪心取

O(N ^ 2) 24分
*/

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