結果

問題 No.1874 Minimum of Sum of Rectangles
ユーザー Kude
提出日時 2022-03-11 23:10:37
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 2,288 bytes
コンパイル時間 2,185 ms
コンパイル使用メモリ 222,496 KB
最終ジャッジ日時 2025-01-28 09:03:40
ジャッジサーバーID
(参考情報)
judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 37 WA * 2
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
namespace {
#pragma GCC diagnostic push
#pragma GCC diagnostic ignored "-Wunused-function"
#include<atcoder/all>
#pragma GCC diagnostic pop
using namespace std;
using namespace atcoder;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; i--)
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
template<class T> bool chmax(T& a, const T& b) { if (a < b) { a = b; return true; } else return false; }
template<class T> bool chmin(T& a, const T& b) { if (b < a) { a = b; return true; } else return false; }
using ll = long long;
using P = pair<int,int>;
using VI = vector<int>;
using VVI = vector<VI>;
using VL = vector<ll>;
using VVL = vector<VL>;
template<class S>
struct value_compression : vector<S> {
bool built = false;
using VS = vector<S>;
using VS::vector;
value_compression(vector<S> v) : vector<S>(move(v)) {}
void build() {
sort(VS::begin(), VS::end());
VS::erase(unique(VS::begin(), VS::end()), VS::end());
built = true;
}
template<class T>
void convert(T first, T last) { for (; first != last; ++first) *first = (*this)(*first); }
int operator()(S x) {
assert(built);
return lower_bound(VS::begin(), VS::end(), x) - VS::begin();
}
void clear() { VS::clear(); built = false; }
};
} int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
VI x(n), y(n);
rep(i, n) cin >> x[i] >> y[i];
value_compression<int> vx = x, vy = y;
vx.build(), vy.build();
const int szx = vx.size(), szy = vy.size();
int lx = vx[0] - 1, rx = vx.back() + 1;
auto score = [&](int a, int b) {
ll res = 0;
rep(i, n) res += (ll)abs(x[i] - a) * abs(y[i] - b);
return res;
};
auto f = [&](int a) {
int ly = vy[0] - 1, ry = vy.back();
while(ry - ly > 2) {
int cy1 = (ly + ry) / 2;
int cy2 = cy1 + 1;
if (score(a, cy1) < score(a, cy2)) ry = cy2;
else ly = cy1;
}
return ly + 1;
};
while(rx - lx > 2) {
int cx1 = (lx + rx) / 2;
int cx2 = cx1 + 1;
int y1 = f(cx1);
int y2 = f(cx2);
if (score(cx1, y1) < score(cx2, y2)) {
rx = cx2;
} else {
lx = cx1;
}
}
int ax = lx + 1;
int ay = f(ax);
ll ans = score(ax, ay);
cout << ans << '\n';
}
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