Submission #916365

#	Time	Username	Problem	Language	Result	Execution time	Memory
916365		jlass	Balloons (CEOI11_bal)	C++17	30 / 100	2056 ms	348 KiB

bal

#include <bits/stdc++.h>
typedef long long ll;
typedef long double ld;
using namespace std;
const ll MOD = 1e9 + 7;
ld eps = 1e-12;
/*
Notes for stack:
store dimensions of balloon, as well as for what distance it is significant for
From bottom to top of the stack, the distance it's significant should be increasing 

Find insert size
Because the significant distance is increasing in size, we pop elements in the stack until the distance is less than the insertion distance
Then, take min of balloon size and stack thing

Use Geometries to find balloon sizing and distances ft. Marvin

*/

ld calc_r(ld dx, ld r1) {
    return dx * dx / 4 / r1;
}

//given that r1 > r2, otherwise we do not care
ld calc_x(ld x1, ld r1, ld x2, ld r2) {
    ld l = x2, r = 1e9 + 1;
    while(r - l > eps) {
        ld m = l + (r - l) / 2;
        ld d1 = calc_r(m - x1, r1);
        ld d2 = calc_r(m - x2, r2);
        if(d1 > d2) {
            l = m;
        } else {
            r = m;
        }
    }
    return l + (r - l) / 2;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout << fixed << setprecision(3);
    int n; cin >> n;
    stack<pair<ld, pair<ld,ld>>> s; //stored as {significant distance, {x, r}}
    for(int i = 0; i < n; i++) {
        ld x, maxr; cin >> x >> maxr;
        while(s.size() && x > s.top().first) {
            s.pop();
        }
        ld r = min(maxr, s.empty() ? maxr : calc_r(x - s.top().second.first, s.top().second.second));
        //current balloon will always be significant for some portion of graph
        while(s.size() && (r > s.top().second.second || calc_x(s.top().second.first, s.top().second.second, x, r) > s.top().first)) {
            s.pop();
        }
        ld dist = (s.empty() ? 1e9 + 5 : calc_x(s.top().second.first, s.top().second.second, x, r));
        s.push({dist, {x,r}});
        cout << r << '\n';
    }

    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 10 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 10
• Subtask 6: 0 / 10
• Subtask 7: 0 / 10
• Subtask 8: 0 / 10
• Subtask 9: 0 / 10
• Subtask 10: 0 / 10