이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "aliens.h"
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 20;
const long long inf = 1e18L + 20;
const long long max_cost = 1e12L + 20;
long long lt[maxn];
long long rt[maxn];
long long over[maxn];
pair<long long, int> dp[maxn];
long long cost;
int n;
struct line {
long long a, b;
int cnt;
line(long long _a, long long _b, int _cnt): a(_a), b(_b), cnt(_cnt) {};
long long eval(long long x) {
return a * x + b;
};
};
long long fdiv(long long x, long long y) {
return (x / y) - ((x ^ y) < 0 && (x > y));
}
long long inter(line L1, line L2) {
return fdiv(L1.b - L2.b, L2.a - L1.a);
}
struct CHT {
vector<line> Q;
int pt = 0;
void add(line L) {
while ((int)Q.size() >= 2 && inter(L, Q.back()) <= inter(Q.back(), Q.end()[-2])) {
Q.pop_back();
}
Q.push_back(L);
}
pair<long long, int> get(long long x) {
pt = min(pt, (int)Q.size());
while (pt + 1 < (int)Q.size() && Q[pt].eval(x) > Q[pt + 1].eval(x)) {
pt++;
}
return {Q[pt].eval(x) + x * x, Q[pt].cnt + 1};
}
};
void solve() {
CHT C;
dp[0] = {0, 0};
C.add(line(2 - lt[1] * 2, lt[1] * lt[1] - lt[1] * 2 + cost + 1, 0));
for (int i = 1; i <= n; i++) {
dp[i] = C.get(rt[i]);
C.add(line(2 - lt[i + 1] * 2, dp[i].first + lt[i + 1] * lt[i + 1] - lt[i + 1] * 2 - over[i] + cost + 1, dp[i].second));
}
}
long long take_photos(int _n, int m, int k, std::vector<int> rows, std::vector<int> cols) {
n = _n;
vector<pair<int, int>> p, q;
for (int i = 0; i < n; i++) {
if (rows[i] > cols[i]) {
swap(rows[i], cols[i]);
}
p.push_back({rows[i], -cols[i]});
}
sort(p.begin(), p.end());
for (auto x: p) {
if (q.empty() || x.second < q.back().second) {
q.push_back(x);
}
}
n = q.size();
k = min(k, n);
for (int i = 0; i < n; i++) {
lt[i + 1] = q[i].first;
rt[i + 1] = -q[i].second;
}
over[0] = 0;
for (int i = 1; i <= n - 1; i++) {
long long sz = max(rt[i] - lt[i + 1] + 1, 0LL);
over[i] = sz * sz;
}
long long cost_l = 0;
long long cost_r = max_cost;
long long res = inf;
while (cost_l <= cost_r) {
cost = (cost_l + cost_r) / 2;
solve();
if (dp[n].second <= k) {
res = dp[n].first - cost * k;
cost_r = cost - 1;
}
else {
cost_l = cost + 1;
}
}
return res;
}
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