Submission #1203022

#	Time	Username	Problem	Language	Result	Execution time	Memory
1203022		zeta7532	Train (APIO24_train)	C++20	0 / 100	50 ms	19008 KiB

train

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll INF = (1LL << 60);

// ---- Persistent Segment Tree for point-add & range-sum ----
struct PST {
    struct Node { int l, r, sum; };
    vector<Node> st;
    int N;
    PST(int _N): N(_N) {
        st.reserve(_N * 20);
        st.push_back({0,0,0});  // dummy node at idx=0
    }
    // build empty tree on [l..r], returns root index
    int build(int l, int r) {
        int id = st.size();
        st.push_back({0,0,0});
        if (l < r) {
            int m = (l + r) >> 1;
            st[id].l = build(l, m);
            st[id].r = build(m+1, r);
        }
        return id;
    }
    // create new version from prev, add v at pos p
    int update(int prev, int l, int r, int p, int v) {
        int id = st.size();
        st.push_back(st[prev]);
        if (l == r) {
            st[id].sum += v;
        } else {
            int m = (l + r) >> 1;
            if (p <= m)
                st[id].l = update(st[prev].l, l, m, p, v);
            else
                st[id].r = update(st[prev].r, m+1, r, p, v);
            st[id].sum = st[st[id].l].sum + st[st[id].r].sum;
        }
        return id;
    }
    // query sum on [ql..qr]
    int query(int id, int l, int r, int ql, int qr) const {
        if (!id || qr < l || r < ql) return 0;
        if (ql <= l && r <= qr) return st[id].sum;
        int m = (l + r) >> 1;
        return query(st[id].l, l, m, ql, qr)
             + query(st[id].r, m+1, r, ql, qr);
    }
};

// solve function to be called by the judge
// signature matches: solve(int n, int m, int w,
//     vector<int> x, vector<int> y,
//     vector<int> a, vector<int> b, vector<int> c,
//     vector<int> l, vector<int> r, vector<int> t)
long long solve(int n, int m, int w,
                vector<int> x, vector<int> y,
                vector<int> a, vector<int> b, vector<int> c,
                vector<int> L, vector<int> R, vector<int> T) {

    // 1) collect all time points for coordinate compression
    vector<int> allT;
    allT.reserve(2*m + 2*w + 1);
    allT.push_back(0);
    for(int i = 0; i < m; i++){
        allT.push_back(a[i]);
        allT.push_back(b[i]);
    }
    for(int i = 0; i < w; i++){
        allT.push_back(L[i]);
        allT.push_back(R[i]);
    }
    sort(allT.begin(), allT.end());
    allT.erase(unique(allT.begin(), allT.end()), allT.end());
    auto comp = [&](int v) {
        return int(lower_bound(allT.begin(), allT.end(), v) - allT.begin());
    };
    int Tsz = allT.size();

    // 2) bucket meals by compressed R
    vector<vector<int>> meals_at_r(Tsz);
    for(int i = 0; i < w; i++){
        int lc = comp(L[i]);
        int rc = comp(R[i]);
        meals_at_r[rc].push_back(lc);
    }

    // 3) build persistent segment tree versions
    PST pst(Tsz);
    vector<int> version(Tsz);
    version[0] = pst.build(0, Tsz-1);
    for(int lc: meals_at_r[0]){
        version[0] = pst.update(version[0], 0, Tsz-1, lc, 1);
    }
    for(int i = 1; i < Tsz; i++){
        version[i] = version[i-1];
        for(int lc: meals_at_r[i]){
            version[i] = pst.update(version[i], 0, Tsz-1, lc, 1);
        }
    }

    // helper to get g(b, a): #meals completely inside (b, a)
    auto get_g = [&](int b, int a) {
        int bc = comp(b), ac = comp(a);
        if (ac == 0) return 0;
        int total = pst.query(version[ac-1], 0, Tsz-1, 0, Tsz-1);
        int bad   = pst.query(version[ac-1], 0, Tsz-1, 0, bc);
        return total - bad;
    };

    // 4) prepare train indices sorted by departure time
    vector<int> idx(m);
    iota(idx.begin(), idx.end(), 0);
    sort(idx.begin(), idx.end(),
         [&](int i, int j){ return a[i] < a[j]; });

    // dp[i]: min cost to finish train i
    vector<ll> dp(m, INF);

    // for each planet, a min-heap of (arrival_time b_j, train_index j)
    vector< priority_queue<pair<int,int>,
        vector<pair<int,int>>, greater<pair<int,int>>> > heap(n);

    // and a deque to maintain candidate js with decision-monotonicity
    vector< deque<int> > dq(n);

    // initial state: planet 0 at time 0, treat j=-1 with dp=-0
    heap[0].push({0, -1});
    auto get_dp = [&](int j){
        return j < 0 ? 0LL : dp[j];
    };
    auto get_b  = [&](int j){
        return j < 0 ? 0    : b[j];
    };

    ll answer = INF;

    // 5) main scan in order of departure time
    for(int _k = 0; _k < m; _k++){
        int i = idx[_k];
        int p = x[i];
        int ai = a[i];

        // step1: move all arriving-by-ai trains from heap[p] into dq[p]
        auto &hp = heap[p];
        auto &dq_p = dq[p];
        while(!hp.empty() && hp.top().first <= ai){
            int j = hp.top().second;
            hp.pop();
            // pop-back by quadrangle-inequality check
            while(dq_p.size() >= 2){
                int j2 = dq_p.back();
                int j1 = dq_p[dq_p.size()-2];
                ll lhs = get_dp(j2)
                       + (ll)get_g(get_b(j2), ai) * T[p];
                ll rhs = get_dp(j1)
                       + (ll)get_g(get_b(j1), ai) * T[p];
                if (lhs >= rhs) dq_p.pop_back();
                else break;
            }
            dq_p.push_back(j);
        }

        // step2: front of dq[p] is best predecessor
        if (!dq_p.empty()){
            int j = dq_p.front();
            dp[i] = get_dp(j)
                  + (ll)get_g(get_b(j), ai) * T[p]
                  + c[i];
            if (y[i] == n-1)
                answer = min(answer, dp[i]);
        }

        // step3: push this train into heap at its arrival planet
        heap[y[i]].push({ b[i], i });
    }

    if (answer >= INF/2) return -1;
    return answer;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 5
• Subtask 3: 0 / 30
• Subtask 4: 0 / 60