Submission #986685

#	Time	Username	Problem	Language	Result	Execution time	Memory
986685		JooDdae	Salesman (IOI09_salesman)	C++17	100 / 100	1085 ms	45560 KiB

salesman

#pragma GCC optimize ("O3")
#pragma GCC target ("avx2")

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

#define mid ((l+r) >> 1)

const int INF = 2e9, N = 5e5+1;

int n, u, d, s, mx[500500], t1[2002002], t2[2002002];
vector<array<int, 2>> v[500500];

void update(int t[], int id, int x, int node = 1, int l = 1, int r = N) {
    if(id < l || r < id) return;
    if(l == r) {
        t[node] = max(t[node], x);
        return;
    }
    if(id <= mid) update(t, id, x, node*2, l, mid);
    else update(t, id, x, node*2+1, mid+1, r);
    t[node] = max(t[node*2], t[node*2+1]);
}

int find(int t[], int nl, int nr, int node = 1, int l = 1, int r = N) {
    if(nr < l || r < nl) return -INF;
    if(nl <= l && r <= nr) return t[node];
    return max(find(t, nl, nr, node*2, l, mid), find(t, nl, nr, node*2+1, mid+1, r));
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    cin >> n >> u >> d >> s;
    for(int i=1;i<=n;i++) {
        int t, x, y; cin >> t >> x >> y;
        v[t].push_back({x, y});
    }

    fill(t1, t1+4*N+1, -INF), fill(t2, t2+4*N+1, -INF);
    fill(mx, mx+N+1, -INF);
    update(t1, s, s*u), update(t2, s, -s*d);
    for(int t=1;t<=N;t++) if(!v[t].empty()) {
        sort(v[t].begin(), v[t].end());

        for(auto [x, y] : v[t]) mx[x] = max(find(t1, 1, x-1) - x*u, x*d + find(t2, x+1, N)) + y;

        for(auto [x, y] : v[t]) {
            update(t1, x, x*u+mx[x]);
            update(t2, x, mx[x]-x*d);
        }

        for(auto [x, y] : v[t]) {
            auto k = find(t1, 1, x-1) - x*u + y;
            mx[x] = max(mx[x], k), update(t1, x, x*u+k);
        }
        reverse(v[t].begin(), v[t].end());
        for(auto [x, y] : v[t]) {
            auto k = x*d + find(t2, x+1, N) + y;
            mx[x] = max(mx[x], k), update(t2, x, k-x*d);
        }

        for(auto [x, y] : v[t]) {
            update(t1, x, x*u+mx[x]);
            update(t2, x, mx[x]-x*d);
        }
    }

    int ans = 0;
    for(int i=1;i<=N;i++) {
        ans = max(ans, find(t1, 1, s-1) - s*u);
        ans = max(ans, s*d + find(t2, s+1, N));
    }
    cout << ans << "\n";
}

• Subtask 1: 100 / 100