/**
* I don't feel like fully solving this, so maybe i'll just score some partials.
*
* [S1] Too trivial
* [S2] Each point lies on the main diagonal. Clearly, no two squares should have
* common squares, which means the points are partitioned into contiguous
* segments. O(n^2 * k) dp kills it.
* [S3] Maybe we can somehow sort the points based on their columns, then apply [S2]?
* Yep, pretty much analagous to [S2]. The key here is that if (2,5), (3,5),
* the square that covers (2,5) must cover (3,5).
*
* Time Complexity: O(n^2 * k)
* Implementation 1
*/
#include <bits/stdc++.h>
#include "aliens.h"
typedef long long ll;
const int N_MAX = 500;
const ll INF = 0x3f3f3f3f3f3f;
inline ll sqr(int k) { return ll(k) * ll(k); }
struct seg_t {
int i, j;
};
ll cover[N_MAX + 1][N_MAX + 1];
ll take_photos(int n, int m, int max_sqr, std::vector<int> row, std::vector<int> col) {
std::vector<seg_t> values(n);
for (int k = 0; k < n; k++)
values[k].i = std::min(row[k], col[k]), values[k].j = std::max(row[k], col[k]);
std::sort(values.begin(), values.end(),
[](const seg_t& s1, const seg_t& s2) {
return s1.j < s2.j || (s1.j == s2.j && s1.i < s2.i);
});
values.insert(values.begin(), seg_t{-1, -1});
cover[0][0] = 0;
for (int squares = 1; squares <= max_sqr; squares++)
cover[0][squares] = INF;
int last_idx = -1;
for (int k = 1, prev_j = -1; k <= n; k++) {
if (prev_j == values[k].j)
continue;
prev_j = values[k].j, last_idx = k;
cover[k][0] = INF;
for (int squares = 1; squares <= max_sqr; squares++) {
cover[k][squares] = INF;
for (int m = k - 1, left = 0x3f3f3f; m >= 0; m--) {
left = std::min(left, values[m + 1].i);
cover[k][squares] = std::min(
cover[k][squares],
cover[m][squares - 1] + sqr(values[k].j - left + 1)
- sqr(std::max(values[m].j - left + 1, 0))
);
}
}
}
ll ans = INF;
for (int squares = 1; squares <= max_sqr; squares++)
ans = std::min(ans, cover[last_idx][squares]);
return ans;
}
