이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "aliens.h"
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")
#include <bits/stdc++.h>
#define ar array
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define ll long long
using namespace std;
template<typename T> bool ckmin(T &a, const T &b) { return a > b ? a = b, 1 : 0; }
template<typename T> bool ckmax(T &a, const T &b) { return a < b ? a = b, 1 : 0; }
const ll INF = 1e18;
const int nax = 1e5 + 5;
ar<ll, 4> d[nax];
long long take_photos(int n, int m, int k, std::vector<int> r, std::vector<int> c) {
vector<ar<int, 2>> a(n);
for (int i = 0; i < n; i++) {
a[i] = {r[i], c[i]};
if (r[i] > c[i]) swap(a[i][0], a[i][1]);
}
sort(all(a));
int mx = -1;
vector<ar<int, 2>> b;
for (int i = 0; i < n; i++) {
int r = i;
while (r+1 < n && a[r+1][0] == a[r][0]) r++;
if (a[r][1] > mx) {
b.push_back({a[r][0], a[r][1]});
mx = a[r][1];
}
i = r;
}
swap(a, b);
n = sz(a);
ll L = 0, R = 1e12, ans = -1;
while (L <= R) {
ll mid = (L+R)/2;
vector<ar<ll, 3>> dp(n+1, {INF, INF, -INF});
dp[0] = {0, 0, 0};
auto sq = [&](int x) { return (ll)x * x; };
auto f = [&](int j) {
return dp[j][0] + sq(a[j][0]) - 2 * a[j][0] - (j ? sq(max(0, a[j-1][1] - a[j][0] + 1)) : 0);
};
int dl = -1, dr = 0;
d[++dl] = {-2 * a[0][0], f(0), 0, 0};
auto query = [&](int p, int pos) {
return d[p][0] * pos + d[p][1];
};
auto inter = [&](const ar<ll, 4>& x1, const ar<ll, 4>& x2) {
return (double)(x2[1] - x1[1]) / (x1[0] - x2[0]);
};
for (int i = 1; i <= n; i++) {
ll C = sq(a[i-1][1]) + 2 * a[i-1][1] + 1 + mid;
int cur = a[i-1][1];
while (dl+1 <= dr && query(dl+1, cur) < query(dl, cur)) dl++;
dp[i][0] = query(dl, cur) + C;
for (int j = dl; j <= dr; j++) {
if (query(j, cur) == query(dl, cur)) {
ckmin(dp[i][1], d[j][2] + 1);
ckmin(dp[i][2], d[j][3] + 1);
}
else break;
}
if (i < n) {
ar<ll, 4> nw{-2 * a[i][0], f(i), dp[i][1], dp[i][2]};
while (dr-dl >= 1 && inter(d[dr-1], nw) <= inter(d[dr-1], d[dr])) dr--;
d[++dr] = nw;
}
}
if (dp[n][1] <= k) {
int u = min(k, int(dp[n][2]));
ans = dp[n][0] - u * mid;
R = mid-1;
}
else L = mid+1;
}
return ans;
}
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