Submission #163230

#TimeUsernameProblemLanguageResultExecution timeMemory
163230qkxwsmNLO (COCI18_nlo)C++14
55 / 110
307 ms65540 KiB
#include <bits/stdc++.h> using namespace std; template<class T, class U> void ckmin(T &a, U b) { if (a > b) a = b; } template<class T, class U> void ckmax(T &a, U b) { if (a < b) a = b; } #define MP make_pair #define PB push_back #define LB lower_bound #define UB upper_bound #define fi first #define se second #define FOR(i, a, b) for (auto i = (a); i < (b); i++) #define FORD(i, a, b) for (auto i = (a) - 1; i >= (b); i--) #define SZ(x) ((int) ((x).size())) #define ALL(x) (x).begin(), (x).end() #define MAXN 1013 typedef long long ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef vector<int> vi; typedef vector<ll> vl; typedef vector<pii> vpi; typedef vector<pll> vpl; int R, C, N; pair<pll, ll> arr[MAXN]; vector<pair<int, short> > event[MAXN * MAXN]; set<int> ok; ll ans; int getsqrt(ll x) { ll res = sqrt(x); while(res * res < x) res++; while(res * res > x) res--; return res; } int32_t main() { ios_base::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(12); cerr << fixed << setprecision(4); cin >> R >> C >> N; FOR(i, 1, N + 1) { int a, b, r; cin >> a >> b >> r; a--; b--; FOR(j, a - r, a + r + 1) { int d = getsqrt(1ll * r * r - (a - j) * (a - j)); event[j].PB({b - d, i}); event[j].PB({b + d + 1, -i}); } } FOR(i, 0, R) { sort(ALL(event[i])); FOR(j, 0, SZ(event[i])) { auto p = event[i][j]; if (!ok.empty()) { ans += (*ok.rbegin()) * (p.fi - event[i][j - 1].fi); } if (p.se < 0) { // cerr << "j = " << j << endl; // cerr << "ERASE " << -p.se << endl; ok.erase(-p.se); } else { // cerr << "INSERT " << p.se << endl; ok.insert(p.se); } } } //find sum(latest day) //the # of poitns that are inside here but inside NONE OF THE FUTURE ONES ans = 1ll * (R * C) * N - ans; cout << ans << '\n'; return 0; }
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