AtCoder ABC 074 D
June 17, 2020
D - Restoring Road Network
解説のように、でも解けるが、座標圧縮 + 二次元累積和で事前に長方形に含まれる点の個数を計算しておくこともできる。
int main() {
i64 N, K; cin >> N >> K;
using P = pair<i64, i64>;
vector<P> point(N);
map<i64, i64> mpX, mpY;
REP(i, N) {
cin >> point[i].first >> point[i].second;
mpX[point[i].first] = 0;
mpY[point[i].second] = 0;
}
vector<i64> xvals;
vector<i64> yvals;
for (auto &p : mpX) {
p.second = xvals.size();
xvals.pb(p.first);
}
for (auto &p : mpY) {
p.second = yvals.size();
yvals.pb(p.first);
}
i64 H = yvals.size() * 2;
i64 W = xvals.size() * 2;
vector<vector<i64>> G(H, vector<i64>(W, 0));
REP(i, N) {
i64 y = mpY[point[i].second] * 2;
i64 x = mpX[point[i].first] * 2;
G[y][x] = 1;
}
vector<vector<i64>> C(H + 1, vector<i64>(W + 1, 0));
REP(y, H) {
REP(x, W) {
C[y + 1][x + 1] = C[y][x + 1] + C[y + 1][x] - C[y][x] + G[y][x];
}
}
i64 mins = numeric_limits<i64>::max();
REP(i, N) REP(j, N) REP(k, N) REP(l, N) {
i64 minx, miny, maxx, maxy;
minx = maxx = point[i].first;
miny = maxy = point[i].second;
auto update = [&](i64 idx) {
chmin(minx, point[idx].first);
chmax(maxx, point[idx].first);
chmin(miny, point[idx].second);
chmax(maxy, point[idx].second);
};
update(j), update(k), update(l);
i64 clx, cly, crx, cry;
clx = mpX[minx] * 2;
cly = mpY[miny] * 2;
crx = mpX[maxx] * 2 + 1;
cry = mpY[maxy] * 2 + 1;
i64 cnt = C[cry][crx] - C[cry][clx] - C[cly][crx] + C[cly][clx];
if (cnt >= K) {
i64 dx = abs(maxx - minx);
i64 dy = abs(maxy - miny);
if (dx * dy > 0) chmin(mins, dx * dy);
}
}
cout << mins << endl;
return 0;
}