#include <bits/stdc++.h>
#include "aliens.h"
#define ll long long
#define int long long
using namespace std;
const int INF = 1e18;
int sqr(int x) {return x * x;}
long long take_photos(int32_t n, int32_t m, int32_t k, vector<int32_t> r, vector<int32_t> c) {
vector<pair<int,int>> tmpcords;
vector<pair<int,int>> cords;
for (int i = 0; i < n; ++i) {
tmpcords.push_back({min(r[i], c[i]), max(r[i], c[i])});
}
sort(tmpcords.begin(), tmpcords.end(),
[](pair<int,int>a,pair<int,int>b)
{return a.first<b.first || (a.first==b.first && a.second>b.second);});
for (auto i : tmpcords) {
if (cords.size() && cords.back().second >= i.second) continue;
cords.push_back(i);
}
n = cords.size();
//cout << "DBG" << endl;
//for (int i =0 ;i < n; ++i) {
//cout << cords[i].first << " " << cords[i].second << "\n";
//}
auto cost = [&](int i, int j) {
// area of photo - overlap
if (i == 0)
return sqr(cords[j].second - cords[i].first + 1);
else
return sqr(cords[j].second - cords[i].first + 1) - max(0LL, sqr(cords[i-1].second - cords[i].first + 1));
};
vector<vector<int>> dp(n+1, vector<int>(k+5, INF));
vector<vector<int>> opt(n+1, vector<int>(k+5, INF));
for (int i = 0; i < n; ++i) {
//opt[i][i] = i; // if you have n == k then optimum is to take picture of every segment
opt[i][1] = 0; // if you have 1 picture optimum is to start at 0
dp[i][1] = cost(0, i);
}
//dp[0][0] = 0;
//for (int i = 0;i <= k; ++i) dp[0][i] = 0;
for (int i = n-1; i >= 0; --i) {
for (int j =2; j <= k; ++j) {
opt[n][j] = n-1;
dp[i][j] = dp[i][j-1];
for (int p = opt[i][j-1]; p <= min(n-2LL, opt[i+1][j]); ++p) {
int cur = dp[p][j-1] + cost(p+1, i);
//int cur = dp[p][j-1] + sqr(cords[i-1].second - cords[p].first + 1);
//if ((p -1 >= 0) && cords[p-1].second >= cords[p].first)
//cur -= sqr(cords[p-1].second - cords[p].first + 1);
if (cur < dp[i][j]) {
opt[i][j] = p;
dp[i][j] = cur;
}
//cout << cur << " ";
//cout << cur << endl;
//= min(dp[i][j], cur);
}
}
}
//for (int i =0; i<=n;++i) {
//for (int j = 0; j<=k;++j) {
//cout << dp[i][j] << " ";
//}
//cout << "\n";
//}
//int ans = INF;
//for (int i = 0; i <= k; ++i) {
//ans = min(ans, dp[n-1][i]);
//}
return dp[n-1][k];
}
