Submission #248634

#	Time	Username	Problem	Language	Result	Execution time	Memory
248634		thecodingwizard	Lightning Conductor (POI11_pio)	C++11	0 / 100	379 ms	16104 KiB

pio

#include <bits/stdc++.h>

using namespace std;

const int inf = 1e9+10;

struct func {
    int h, i, l, r;
};

// returns when right part of func a intersects func b
int intersectRight(func a, func b) {
    if (a.h >= b.h) return inf;
    long long d = b.h-a.h;
    long double x = (double)(b.i-a.i-d*d)/2/d;
    long long ans = ceil(x*x)+b.i;
    if (ans > inf) return inf;
    return ans;
}

int n; 
int mx[500000];
int mx2[500000];
int H[500000];

void solveRight(int mx[]) {
    stack<func> s;
    for (int i = 0; i < n; i++) {
        func f = {H[i], i, i, n-1};
        while (!s.empty()) {
            func f2 = s.top();
            int intersect = intersectRight(f2, f);
            //cout << f2.i << " intersects " << f.i << " at " << intersect << endl;
            if (intersect < i) {
                s.top().r = i-1;
                break;
            }
            if (intersect <= f2.l) s.pop();
            else {
                if (intersect - 1 <= n-1) s.top().r = intersect - 1;
                f.l = intersect;
                break;
            }
        }
        if (f.l <= n-1) s.push(f);
    }
    while (!s.empty()) {
        func f = s.top(); s.pop();
        //cout << f.i << " is max from " << f.l << " to " << f.r << endl;
        for (int i = f.l; i <= f.r; i++) mx[i] = max(mx[i], f.h+(int)ceil(sqrt(i-f.i)));
    }
    for (int i = 0; i < n; i++) {
        //cout << "mx[" << i << "] = " << mx[i] << endl;
    }
}

int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> H[i]; mx[i] = H[i];
    }
    solveRight(mx);
    for (int i = 0; i < n/2; i++) {
        swap(H[i], H[n-i-1]);
    }
    solveRight(mx2);
    for (int i = 0; i < n; i++) {
        mx[i] = max(mx[i], mx2[n-1-i]);
    }
    for (int i = 0; i < n/2; i++) {
        swap(H[i], H[n-i-1]);
    }
    for (int i = 0; i < n; i++) {
        cout << mx[i]-H[i] << "\n";
    }
    return 0;
}

• Subtask 1: 0 / 9
• Subtask 2: 0 / 9
• Subtask 3: 0 / 9
• Subtask 4: 0 / 9
• Subtask 5: 0 / 9
• Subtask 6: 0 / 9
• Subtask 7: 0 / 9
• Subtask 8: 0 / 9
• Subtask 9: 0 / 9
• Subtask 10: 0 / 9
• Subtask 11: 0 / 10