Submission #759220

#	Time	Username	Problem	Language	Result	Execution time	Memory
759220		PurpleCrayon	Magic Tree (CEOI19_magictree)	C++17	47 / 100	2082 ms	35824 KiB

magictree

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define sz(v) int(v.size())
#define ar array
const int N = 1e5+10, MOD = 1e9+7;
const ll INF = 1e18+10;

struct S {
    // i must be >= the max
    map<int, ll> mp;
    S() {
        mp[0] = 0;
    }

    void add_range(int l, int r, ll x) {
        for (auto& [k, v] : mp) {
            if (l <= k && k <= r) {
                v += x;
            }
        }
    }

    void merge(S& nxt) {
        if (sz(mp) < sz(nxt.mp)) {
            swap(mp, nxt.mp);
        }

        for (auto [k, v] : nxt.mp) {
            ll use = (--mp.upper_bound(k))->second;
            mp[k] = use;
        }

        for (auto one = nxt.mp.begin(); one != nxt.mp.end(); one++) {
            auto two = next(one);
            if (two == nxt.mp.end()) {
                // one->first, to the end
                add_range(one->first, MOD, one->second);
            } else {
                // one->first, two->first-1
                add_range(one->first, two->first - 1, one->second);
            }
        }
    }

    void add(int t, int x) {
        ll cur = (--mp.upper_bound(t))->second + x;
        auto it = mp.lower_bound(t);
        while (it != mp.end() && it->second <= cur) {
            mp.erase(it);
            it = mp.lower_bound(t);
        }

        mp[t] = cur;
    }

    ll get_ans() {
        return mp.rbegin()->second;
    }
} s[N];

int n, m, k;
vector<int> adj[N];
int t[N], a[N];

// value of ancestor has to be >= value of any thing in the subtree
void dfs(int c, int p) {
    s[c] = S();
    for (int nxt : adj[c]) if (nxt != p) {
        dfs(nxt, c);
        s[c].merge(s[nxt]);
    }

    if (t[c] != -1) s[c].add(t[c], a[c]);
}

void solve() {
    cin >> n >> m >> k;
    for (int i = 1; i < n; i++) {
        int p; cin >> p, --p;
        adj[p].push_back(i);
    }

    memset(t, -1, sizeof(t));
    memset(a, -1, sizeof(a));

    while (m--) {
        int c, i, x; cin >> c >> i >> x, --c;
        t[c] = i, a[c] = x;
    }

    dfs(0, -1);
    cout << s[0].get_ans() << '\n';
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    int T = 1;
    // cin >> T;
    while (T--) solve();
}

• Subtask 1: 6 / 6
• Subtask 2: 0 / 3
• Subtask 3: 0 / 11
• Subtask 4: 12 / 12
• Subtask 5: 16 / 16
• Subtask 6: 13 / 13
• Subtask 7: 0 / 22
• Subtask 8: 0 / 17