#include "aliens.h"
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
/*
Cells (r, c) and (c, r) are equivalent.
(a, b) = (min(r, c), max(r, c))
A square (p, q) covers point (a, b) if and only if p <= a and b <= q
If (a1, b1) and (a2, b2) are points and a1 <= a2 and b2 <= b1, then (a2, b2) is irrelevant.
So, for any (a1, b1) and (a2, b2) a1 < a2 <=> b1 < b2
When checking for overlapping areas, we only have to consider the last square.
Let points be sorted (a[1], b[1]) < (a[2], b[2]) < .... < (a[x], b[x]).
Let dp[i][k] be the minimum area required to cover the first i points using k squares.
dp[0][k] = 0
dp[i][k] = min{dp[j][k-1] + (b[i] - a[j] + 1)^2 - max(0, b[j] - a[j] + 1)^2 | j=1..i-1}
A point (a, b) requires a square of endpoints (a, a) and (b, b) (side = b-a+1)
*/
vector<int> R, C;
vector<long long> a(1), b(1);
long long sq(long long x)
{
return x*x;
}
const long long INF = 1'000'000'000'001;
long long take_photos(int n, int m, int k, vector<int> r, vector<int> c)
{
cerr << '\n';
R = r;
C = c;
int I[n];
for(int i = 0; i < n; i++)
I[i] = i;
sort(I, I+n, [] (int x, int y)
{
if(min(R[x], C[x]) == min(R[y], C[y]))
return max(R[x], C[x]) > max(R[y], C[y]);
return min(R[x], C[x]) < min(R[y], C[y]);
});
int maxb = -1;
for(int i:I)
{
// cerr << i << ' ';
if(maxb < max(r[i], c[i]))
{
maxb = max(r[i], c[i]);
a.push_back(min(r[i], c[i]));
b.push_back(max(r[i], c[i]));
}
}
n = a.size() - 1;
// for(int i = 1; i <= n; i++) cerr << a[i] << ' ' << b[i] << '\n';
k = min(k, n);
long long dp[n+1][k+1];
for(int j = 0; j <= k; j++) dp[0][j] = 0;
for(int i = 1; i <= n; i++)
{
// cerr << "i = " << i << ": \n";
dp[i][0] = INF;
dp[i][1] = sq(b[i] - a[1] + 1);
// cerr << dp[i][1] << ' ';
for(int x = 2; x <= k; x++)
{
dp[i][x] = INF;
for(int j = 1; j < i; j++)
{
// cerr << i << ' ' << x << ' ' << j << ' ';
// cerr << dp[j][x-1] << ' ' << sq(b[i] - a[j+1] + 1) << ' ' << sq(max(0LL, b[j] - a[i] + 1)) << '\n';
if(b[j] - a[j+1] + 1 >= 1)
dp[i][x] = min(dp[i][x], dp[j][x-1] + sq(b[i]-a[j+1]+1) - sq(b[j]-a[j+1]+1));
else
dp[i][x] = min(dp[i][x], dp[j][x-1] + sq(b[i]-a[j+1]+1));
}
// cerr << dp[i][x] << ' ';
}
// cerr << '\n';
}
return dp[n][k];
}
