Submission #59181

#	Time	Username	Problem	Language	Result	Execution time	Memory
59181		Benq	Mobile (BOI12_mobile)	C++11	60 / 100	325 ms	132096 KiB

mobile

#include <bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;

#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()

const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 100001;

int N,L;
vpi v;
vector<ld> tmp;

ld perp(pd x, pd y) {
    return (y.f*y.f+y.s*y.s-x.f*x.f-x.s*x.s)/(2*y.f-2*x.f);
}

ld dist(pd a, pd b) {
    return sqrt(pow(a.f-b.f,2)+pow(a.s-b.s,2));
}

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> N >> L;
    F0R(i,N) {
        int x,y; cin >> x >> y; y = abs(y);
        if (sz(v) && v.back().f == x) {
            if (y >= v.back().s) continue;
            else {
                v.pop_back();
                if (sz(tmp)) tmp.pop_back();
            }
        }
        while (sz(tmp) && tmp.back() >= perp(v.back(),{x,y})) {
            v.pop_back();
            tmp.pop_back();
        }
        if (sz(v)) tmp.pb(perp(v.back(),{x,y}));
        v.pb({x,y});
    }
    ld ans = 0;
    F0R(i,sz(v)) {
        ld mn = (i == 0 ? -INF : tmp[i-1]);
        ld mx = (i == sz(v)-1 ? INF : tmp[i]);
        if (mx < 0 || mn > L) continue;
        mn = max(mn,(ld)0), mx = min(mx,(ld)L);
        ans = max(ans,dist(v[i],{mn,0}));
        ans = max(ans,dist(v[i],{mx,0}));
    }
    cout << fixed << setprecision(6) << ans;
}

/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( ) 
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/

• Subtask 1: 4 / 4
• Subtask 2: 4 / 4
• Subtask 3: 4 / 4
• Subtask 4: 4 / 4
• Subtask 5: 4 / 4
• Subtask 6: 5 / 5
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• Subtask 10: 5 / 5
• Subtask 11: 5 / 5
• Subtask 12: 5 / 5
• Subtask 13: 5 / 5
• Subtask 14: 0 / 5
• Subtask 15: 0 / 5
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• Subtask 21: 0 / 5