Problem:
Score: 500 points

Problem Statement

New AtCoder City has an infinite grid of streets, as follows:

At the center of the city stands a clock tower. Let (0, 0) be the coordinates of this point.

A straight street, which we will call East-West Main Street, runs east-west and passes the clock tower. It corresponds to the x-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to East-West Main Street, with a distance of 1 between them. They correspond to the lines \ldots, y = -2, y = -1, y = 1, y = 2, \ldots in the two-dimensional coordinate plane.

A straight street, which we will call North-South Main Street, runs north-south and passes the clock tower. It corresponds to the y-axis in the two-dimensional coordinate plane.

There are also other infinitely many streets parallel to North-South Main Street, with a distance of 1 between them. They correspond to the lines \ldots, x = -2, x = -1, x = 1, x = 2, \ldots in the two-dimensional coordinate plane.

There are N residential areas in New AtCoder City. The i-th area is located at the intersection with the coordinates (X_i, Y_i) and has a population of P_i. Each citizen in the city lives in one of these areas.

The city currently has only two railroads, stretching infinitely, one along East-West Main Street and the other along North-South Main Street.

M-kun, the mayor, thinks that they are not enough for the commuters, so he decides to choose K streets and build a railroad stretching infinitely along each of those streets.

Let the walking distance of each citizen be the distance from his/her residential area to the nearest railroad.

M-kun wants to build railroads so that the sum of the walking distances of all citizens, S, is minimized.

For each K = 0, 1, 2, \dots, N, what is the minimum possible value of S after building railroads?

Constraints

1 \leq N \leq 15

-10 \ 000 \leq X_i \leq 10 \ 000

-10 \ 000 \leq Y_i \leq 10 \ 000

1 \leq P_i \leq 1 \ 000 \ 000

The locations of the N residential areas, (X_i, Y_i), are all distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X_1 Y_1 P_1
X_2 Y_2 P_2
: : :
X_N Y_N P_N

Output

Print the answer in N+1 lines.

The i-th line (i = 1, \ldots, N+1) should contain the minimum possible value of S after building railroads for the case K = i-1.

Sample Input 1

3
1 2 300
3 3 600
1 4 800

Sample Output 1

2900
900
0
0

When K = 0, the residents of Area 1, 2, and 3 have to walk the distances of 1, 3, and 1, respectively, to reach a railroad.

Thus, the sum of the walking distances of all citizens, S, is 1 \times 300 + 3 \times 600 + 1 \times 800 = 2900.

When K = 1, if we build a railroad along the street corresponding to the line y = 4 in the coordinate plane, the walking distances of the citizens of Area 1, 2, and 3 become 1, 1, and 0, respectively.

Then, S = 1 \times 300 + 1 \times 600 + 0 \times 800 = 900.

We have many other options for where we build the railroad, but none of them makes S less than 900.

When K = 2, if we build a railroad along the street corresponding to the lines x = 1 and x = 3 in the coordinate plane, all citizens can reach a railroad with the walking distance of 0, so S = 0. We can also have S = 0 when K = 3.

The figure below shows the optimal way to build railroads for the cases K = 0, 1, 2.

The street painted blue represents the roads along which we build railroads.

Sample Input 2

5
3 5 400
5 3 700
5 5 1000
5 7 700
7 5 400

Sample Output 2

13800
1600
0
0
0
0

The figure below shows the optimal way to build railroads for the cases K = 1, 2.

Sample Input 3

6
2 5 1000
5 2 1100
5 5 1700
-2 -5 900
-5 -2 600
-5 -5 2200

Sample Output 3

26700
13900
3200
1200
0
0
0

The figure below shows the optimal way to build railroads for the case K = 3.

Sample Input 4

8
2 2 286017
3 1 262355
2 -2 213815
1 -3 224435
-2 -2 136860
-3 -1 239338
-2 2 217647
-1 3 141903

Sample Output 4

2576709
1569381
868031
605676
366338
141903
0
0
0

The figure below shows the optimal way to build railroads for the case K = 4.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
 
public class Main {
 
    void solve() throws IOException {
        n = ni();
        x = new int[n];
        y = new int[n];
        p = new int[n];
        ArrayList<ArrayList<Integer>> towns = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            x[i] = ni();
            y[i] = ni();
            p[i] = ni();
            towns.add(new ArrayList<>(Arrays.asList(x[i],y[i],p[i])));
        }
        Collections.sort(towns, (xx,yy) -> (yy.get(2).compareTo(xx.get(2))));
        ArrayList<Integer> xrail = new ArrayList<>();
        xrail.add(0);
        ArrayList<Integer> yrail = new ArrayList<>();
        yrail.add(0);

        long score = score(xrail, yrail);
        out.println(score);
        for (int i = 1; i <= n; i++) {
            long minscore = Long.MAX_VALUE;
            for (int j = 0; j < (1<<i); j++) {
                ArrayList<Integer> xr = new ArrayList<>();
                ArrayList<Integer> yr = new ArrayList<>();
                xr.add(0);
                yr.add(0);
                for (int k = 0; k < i; k++) {
                    if((j>>(k-1)) ==0){
                        yr.add(towns.get(k).get(1));
                        // xr.add(0);
                    }else{
                        xr.add(towns.get(k).get(0));
                        // yr.add(0);
                    }
                }
                minscore = Math.min(minscore, score(xr,yr));
            }
            out.println(minscore);
        }
    }
    int n;
    int[] x;
    int[] y;
    int[] p;
    long score(ArrayList<Integer>xrail, ArrayList<Integer> yrail){
        long score = 0;
        for (int j = 0; j < n; j++) {
            long mindis = 30000;
            for (int k = 0; k < xrail.size(); k++) {
                mindis = Math.min(mindis,Math.abs(x[j]-xrail.get(k)));
            }
            for (int k = 0; k < yrail.size(); k++) {
                mindis = Math.min(mindis,Math.abs(y[j]-yrail.get(k)));
            }
            score += mindis * p[j];
        }
        return score;
    }
    final int mod = 1000000007;
    final BigInteger MOD = BigInteger.valueOf(mod);
    int mul(int x, int y){
        int val = (int)((x * 1L * y) % mod);
        return val>=0 ? val : val+mod;
    }
    int add(int x, int y) {
        x += y;
        if(x < 0) x += mod;
        if(x>=mod) x -= mod;
        return x;
    }
    int sub(int x, int y){
        x = add(x,mod-y);
        if(x < 0) x += mod;
        if(x>=mod) x -= mod;
        return x;
    }
    String ns() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine(), " ");
        }
        return tok.nextToken();
    }
 
    int ni() throws IOException {
        return Integer.parseInt(ns());
    }
 
    long nl() throws IOException {
        return Long.parseLong(ns());
    }
 
    double nd() throws IOException {
        return Double.parseDouble(ns());
    }
 
    String[] nsa(int n) throws IOException {
        String[] res = new String[n];
        for (int i = 0; i < n; i++) {
            res[i] = ns();
        }
        return res;
    }
 
    int[] nia(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = ni();
        }
        return res;
    }
 
    long[] nla(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nl();
        }
        return res;
    }
 
    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;
 
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        tok = new StringTokenizer("");
        Main main = new Main();
        main.solve();
        out.close();
    }
}