Submission #492885

#	Time	Username	Problem	Language	Result	Execution time	Memory
492885		sberens	Balloons (CEOI11_bal)	C++17	100 / 100	230 ms	11716 KiB

bal

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

using namespace __gnu_pbds;
template<typename K> using hset = gp_hash_table<K, null_type>;
template<typename K, typename V> using hmap = gp_hash_table<K, V>;


using namespace std;

#define all(x) (x).begin(), (x).end()
#define pb push_back
#define eb emplace_back
#define smax(x, y) (x = max(x, y))
#define smin(x, y) (x = min(x, y))

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define F0R(i, a) FOR(i,0,a)
#define ROF(i, a, b) for (int i = (b)-1; i >= (a); --i)
#define R0F(i, a) ROF(i,0,a)


using ll = long long;
using ld = long double;

template<typename T>
using vv = vector<vector<T>>;

using vi = vector<int>;
using ii = array<int, 2>;
using vii = vector<ii>;
using vvi = vv<int>;

using vll = vector<ll>;
using l2 = array<ll, 2>;
using vl2 = vector<l2>;
using vvll = vv<ll>;

template<typename T>
using minq = priority_queue<T, vector<T>, greater<T>>;
template<typename T>
using maxq = priority_queue<T>;

const ll M = 1000000007;
const ll infll = M * M;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    vector<array<ld, 2>> b(n);
    F0R(i, n) cin >> b[i][0] >> b[i][1];
    // if the current circle is taller than some circles before it, then a new circle cannot intersect them
    // so use a stack somehow to keep track of the circles that we can intersect
    // obviously? monotonically decreasing

    // now say we have this stack of circles with (x, r) in decreasing r (r is how much they were actually filled)
    // and we want to insert a new circle (x', r')
    // for each circle c in the stack, the radius of the new circle if it is to touch c, r' will be ((x - x')^2 + r^2) / (2 * r)
    // now we go down the stack and while r' > c.r
        // smin(r', min(r_for_tangent, max_r))
    stack<int> s;
    F0R(i, n) {
        auto &[xx, rr] = b[i];
        while (!s.empty()) {
            auto [x, r] = b[s.top()];
            ld tanr = pow(xx - x, 2)/ (4.l * r);
            smin(rr, tanr);
            if (rr >= r) s.pop();
            else break;
        }
        s.push(i);
    }
    cout << fixed << setprecision(3);
    for (auto [x, r] : b) {
        cout << r << '\n';
    }
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 10 / 10
• Subtask 5: 10 / 10
• Subtask 6: 10 / 10
• Subtask 7: 10 / 10
• Subtask 8: 10 / 10
• Subtask 9: 10 / 10
• Subtask 10: 10 / 10