Submission #1016047

#	Time	Username	Problem	Language	Result	Execution time	Memory
1016047		ByeWorld	Balloons (CEOI11_bal)	C++14	50 / 100	2069 ms	4800 KiB

bal

#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
#define ll long long
#define int long long
#define pb push_back
#define fi first
#define se second
#define lf (id<<1)
#define rg ((id<<1)|1)
#define md ((l+r)/2)
#define ld long double
using namespace std;
typedef pair<int,int> pii;
typedef pair<pii, int> ipii;
const int MAXN = 2e5+10;
const int MAXA = 1e6+10;
const int INF = 1e9+10;
// const int INF2 = 1e18+10;
const int LOG = 19;
const int MOD = 1e9+7;
const int SQRT = 450;
const vector<int> NOL = {};
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const vector<int> dx = {1, -1, 0, 0};
const vector<int> dy = {0, 0, 1, -1};
void chmn(ld &a, ld b){ a = min(a, b); }
void chmx(ld &a, ld b){ a = max(a, b); }

int n;
int x[MAXN], mx[MAXN];
ld r[MAXN];

ld dis(int j, int i){
    return (ld)((x[j]-x[i])*(x[j]-x[i]) / (4*r[j]));
}
signed main(){
    cin >> n;
    for(int i=1; i<=n; i++){
        cin >> x[i] >> mx[i];
    }
    vector <int> stac; stac.pb(1);
    r[1] = mx[1];
    for(int i=2; i<=n; i++){
        ld mn = (ld)mx[i];
        if(stac.size() <= 2){
            for(auto j : stac){
                chmn(mn, dis(j, i));
            }
        } else {
            while(stac.size()>=2 && 
                dis(stac[stac.size()-2], i) < dis(stac.back(), i)) stac.pop_back();
            // if(stac.size()==1) chmn(mn, dis(stac.back(), i));
            // else {
            //     chmn(mn, dis(stac.back(), i));
            //     chmn(mn, dis(stac[stac.size()-2], i));
            // }
            int idx = -1;
            for(auto j : stac){
                if(mn > dis(j, i)) idx = j;
                chmn(mn, dis(j, i));
            }
            if(idx != -1){
                while(stac.back() != idx) stac.pop_back();
            }
        }
        r[i] = mn; stac.pb(i);
    }
    for(int i=1; i<=n; i++) cout << fixed << setprecision(3) << r[i] << '\n';
}

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