Submission #474246

#	Time	Username	Problem	Language	Result	Execution time	Memory
474246		hoanghq2004	Magic Tree (CEOI19_magictree)	C++14	100 / 100	141 ms	32680 KiB

magictree

/*
    Central European Olympiad in Informatics 2019 (Day 2)

    Using dynamic programming: dp[u][t] is the maximum of fruits can be harvested at time point earlier than or equal t
    Formula: dp[u][t] = sum(dp[v][t]) (if t < t[u])
                      = max(sum(dp[v][t]), sum(dp[v][t[u]]) + w[u]) otherwise
    So easy to find that: dp[u][t] is an non-decreasing array with t = 0...k and the number of distinct values will smaller than or equal the number of vertex in the subtree rooted u + 1
and the values only can be increase if t = t[v]
    Optimize:
        - map <int, long long> opt[u] contains all values of dp[u][i] - dp[u][i - 1]
        - merge small to large so the complexity is O(n * log^2(n))
        - for each subtree rooted u, dp[u][t[u]] += w[u], so opt[u][t[u]] += w and the values after t[u] have to be increase that: dp[u][t] = max(dp[u][t], w[u]).
          here, define cur = w[u] - dp[u][t] with t = t[u]...
          while opt[u][t] < w[u], dp[u][t] is smaller than w[u], so we should assign dp[u][t] = w[u]
          subtract opt[u][t] from cur and assign opt[u][t] = 0 (erase from map)
          in other case, dp[u][t] is larger than w[u] then all values from t shouldn't be changed, subtract w[u] from opt[u][t]
*/

#include <bits/stdc++.h>

using namespace std;

int n, m, k;
vector <int> e[100010];
int w[100010], t[100010];
map <int, long long> opt[100010];
long long res;

void dfs(int u) {
    for (auto v: e[u]) {
        dfs(v);
        if (opt[u].size() < opt[v].size()) opt[u].swap(opt[v]);
        for (auto it: opt[v]) opt[u][it.first] += it.second;
        opt[v].clear();
    }
    if (w[u] == 0) return;
    opt[u][t[u]] += w[u];
    auto it = opt[u].upper_bound(t[u]);
    while (it != opt[u].end()) {
        if (w[u] >= it -> second) {
            w[u] -= it -> second;
            opt[u].erase(it++);
        }
        else {
            it -> second -= w[u];
            break;
        }
    }
}

int main() {
    ios_base :: sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    cin >> n >> m >> k;
    for (int i = 2; i <= n; ++i) {
        int p;
        cin >> p;
        e[p].push_back(i);
    }
    while (m--) {
        int u;
        cin >> u;
        cin >> t[u] >> w[u];
    }
    dfs(1);
    for (auto it: opt[1]) res += it.second;
    cout << res;
}

• Subtask 1: 6 / 6
• Subtask 2: 3 / 3
• Subtask 3: 11 / 11
• Subtask 4: 12 / 12
• Subtask 5: 16 / 16
• Subtask 6: 13 / 13
• Subtask 7: 22 / 22
• Subtask 8: 17 / 17