Submission #399365

#	Time	Username	Problem	Language	Result	Execution time	Memory
399365		12tqian	Worst Reporter 4 (JOI21_worst_reporter4)	C++17	14 / 100	2082 ms	5080 KiB

worst_reporter2

#include <bits/stdc++.h>

using namespace std;

#define f1r(i, a, b) for (int i = a; i < b; ++i)
#define f0r(i, a) f1r(i, 0, a)
#define each(t, a) for (auto& t : a)

#define mp make_pair
#define f first
#define s second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;

template <class T> bool ckmin(T& a, const T& b) { return b < a ? a = b, 1 : 0; }
template <class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }

template <class T> int get_pos(vector<T> v, T x) {
    return lower_bound(all(v), x) - v.begin();
}

const int N = 2e5 + 5;

struct DSU {
    vi e;
    
    void init(int n) {
        e = vi(n, -1);
    }

    int get(int x) {
        return e[x] < 0 ? x : e[x] = get(e[x]);
    }

    int size(int x) {
        return -e[get(x)];
    }

    bool unite(int x, int y) {
        x = get(x), y = get(y);
        if (x == y) return false;
        if (e[x] > e[y]) swap(x, y);
        e[x] += e[y]; e[y] = x;
        return true;
    }
};


int main() {
    cin.tie(0)->sync_with_stdio(0);
    int n; cin >> n;
    vi a(n);
    vi h(n);
    vl c(n);
    vector<bool> vis(n);
    vector<bool> cyc(n);
    vi tmp;
    f0r(i, n) {
        cin >> a[i] >> h[i] >> c[i];
        --a[i];
        tmp.pb(h[i]);
    }
    sort(all(tmp));
    tmp.erase(unique(all(tmp)), tmp.end());
    f0r(i, n) h[i] = get_pos(tmp, h[i]);
    f0r(i, n) h[i] = sz(tmp) - 1 - h[i];
    DSU D; D.init(n);
    // if (n == 5001) return 0;
    int cnt = 0;
    f0r(i, n) {
        if (vis[i]) continue;
        int cur = i;
        vi path;
        while (!vis[cur]) {
            vis[cur] = true;
            path.pb(cur);
            cur = a[cur];
            cnt++;
            assert(cnt <= 1e6);
        }
        int loc = -1;
        f0r(i, sz(path)) {
            if (path[i] == cur) {
                loc = i;
                break;
            }
        }
        if (loc == -1) continue;
        f1r(i, loc, sz(path)) {
            cyc[path[i]] = true;
            D.unite(cur, path[i]);
        }
    }
    vi v;
    f0r(i, n) v.pb(D.get(i));
    sort(all(v));
    v.erase(unique(all(v)), v.end());
    vi id(n);
    f0r(i, n) id[i] = get_pos(v, D.get(i));
    int sz = sz(v);
    vector<map<int, ll>> dp(sz);
    vector<vi> comps(sz);
    f0r(i, n) {
        comps[id[i]].pb(i);
    }
    vi roots;
    each(x, v) {
        if (cyc[comps[get_pos(v, x)][0]]) {
            roots.pb(get_pos(v, x));
        }
    }
    vector<vi> g(sz);
    f0r(i, n) {
        int x = id[i];
        int y = id[a[i]];
        if (x == y) continue;
        g[x].pb(y);
        g[y].pb(x);
    }
    ll ans = 0;
    each(r, roots) {
        function<void(int, int)> dfs = [&](int u, int p) {
            each(v, g[u]) {
                if (v == p) continue;
                dfs(v, u);
                if (sz(dp[v]) > sz(dp[u])) swap(dp[u], dp[v]);
                each(x, dp[v]) {
                    dp[u][x.f] += x.s;
                }
            } 
            // add the root
            auto& st = dp[u];
            if (u == r) {
                auto it = st.begin();
                while (it != st.end()) {
                    if (it != st.begin()) {
                        it->s += prev(it)->s;
                    }
                    it = next(it);
                }
                map<int, ll> cost;
                each(vert, comps[u]) {
                    cost[h[vert]] += c[vert];
                }
                ll mx = 0;
                if (sz(st)) mx = st.rbegin()->s;
                each(vert, cost) {
                    int d = vert.f;
                    ll v = vert.s;
                    it = st.upper_bound(d);
                    if (it != st.begin()) {
                        ckmax(mx, prev(it)->s + v);
                    }
                    ckmax(mx, v);
                }
                ans += mx;
            } else {
                int vert = comps[u][0];
                int d = h[vert];
                ll v = c[vert];
                st[d] += v;
                auto it = st.upper_bound(d);
                while (it != st.end() && v) {
                    ll rem = min(v, it->s);
                    it->s -= rem;
                    v -= rem;
                    if (it->s == 0) {
                        it = next(it);
                        st.erase(prev(it));
                    }
                }
            }
        };
        dfs(r, -1);
    }
    ll sum = 0;
    f0r(i, n) sum += c[i];
    ans = sum - ans;
    cout << ans << '\n';
    return 0;
}

/**
 * goes to a[i]
 * maximize the cost if decreasing subsequences up the tree
 * let's consider the cycle at the very end
 * then there is a cost associated with each possible value at the top
 * and a cost associated with nothing
 * first consider the case of fixed
 * things decrease from the root
 * this means that in increasing order it's better
 */

• Subtask 1: 14 / 14
• Subtask 2: 0 / 65
• Subtask 3: 0 / 21