/*
首先考虑点在直线的两边效果一样 于是转移到一边
之后发现当我们覆盖某些点时，有其他的一些点一定会被覆盖  我们找出所有必须覆盖的点
之后我们发现我们找到的这些点 将其按照x递增排序 那么y也是递增的
然后我们 可以得到转移方程了 令dp[i][j]表示前i个都覆盖到了，用了j个正方形的最小花费然后我们考虑转移时的枚举dp[i][j] = min(dp[i][k] + (x[i] - y[k + 1] + 1) ^ 2;这个是显然NKN的， 斜率优化后NK， wqs二分后Nlogm *///#include "aliens.h"
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
using namespace std;
#define M 100010
#define ll long long
#define inf 0x3f3f3f3f
struct Note{ll x, y;bool operator < (const Note &b) const{return this->x == b.x ? this->y > b.y : this->x < b.x;}
}note[M], p[M];
int tot,q[M],h,t, sm[M];
ll dp[M], ans;double X(int x) {return p[x + 1].y;}
double Y(int x) {return dp[x] - 2 * p[x + 1].y + p[x + 1].y * p[x + 1].y;}
double slope(int x, int y) {return (Y(x) - Y(y)) / (X(x) - X(y));} int solve(ll x)
{h = 1, t = 1, q[1] = 0;for(int i = 1; i <= tot; i++){while(h < t && 2.0 * p[i].x > slope(q[h], q[h + 1])) h++;dp[i] = dp[q[h]] + (p[i].x - p[q[h] + 1].y + 1) * (p[i].x - p[q[h] + 1].y + 1) + x;sm[i] = sm[q[h]] + 1;if(i == tot) break;if(p[i + 1].y <= p[i].x) dp[i] -= (p[i].x - p[i + 1].y + 1) * (p[i].x - p[i + 1].y + 1);while(h < t && slope(q[t], q[t - 1]) > slope(q[t], i)) t--;q[++t] = i; }return sm[tot];
}long long take_photos(int n, int m, int k, std::vector<int> r, std::vector<int> c) {for(int i = 0; i < n; i++) {note[i].y = r[i], note[i].x = c[i];if(note[i].x < note[i].y) swap(note[i].x, note[i].y);}sort(note, note + n);for(int i = 1; i < n; i++)    {while(h <= t && note[i].y <= note[q[t]].y ) t--;q[++t] = i;}for(int i = h; i <= t; i++){tot++;p[tot].x = note[q[i]].x, p[tot].y = note[q[i]].y;}if((ans = solve(0)) <= k) {return dp[tot];}ll L = 0, R = 1ll * m * m + 10;while(L <= R){ll mid = (L + R) >> 1;if(solve(mid) <= k) R = mid - 1, ans = dp[tot];else L = mid + 1;}
//    if(ans - 1ll * R * k - k == 568367396612){solve(107455936237); return sm[tot];}return ans - 1ll * R * k - k;
}
/*
5 7 2
0 4 4 4 4
3 4 6 5 62 6 2
1 4
4 14 4 4
1 0 2 2
3 1 1 21 3
0 1
2 1
2 2
*/