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.



Sample input:
3
1 2 300
3 3 600
1 4 800


Sample output:
2900
900
0
0

Submission status: Time Limit Exceeded

Solution code:
import sys
input = sys.stdin.buffer.readline
from itertools import product
INF = 10**18

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: max(abs(xyp[0]), abs(xyp[1])) * xyp[2], reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
import sys
input = sys.stdin.buffer.readline
from itertools import product
INF = 10**18

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=False)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
import sys
input = sys.stdin.buffer.readline
from itertools import product
INF = 10**18

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
INF = 10**18

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
INF = 10**18
N = int(input())

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

def sol():
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
from random import randint, seed
from time import time
seed(time())

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=(randint(0, 3) != 0))

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
from random import randint

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=(randint(0, 3) != 0))

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
from random import randint

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=randint(0, 3)==0)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product
from random import randint

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=randint(0, 1))

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: xyp[2], reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    XYP.sort(key=lambda xyp: abs(xyp[0] * xyp[1] * xyp[2]), reverse=True)

    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    P = [p for _, _, p in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    P = [p for _, _, p in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for i in range(N):
            cost += min(cx[i], cy[i])
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    P = [p for _, _, p in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost = 0
        for s in map(min, zip(cx, cy)):
            cost += s
            if cost >= ans[M]:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

def sol():
    N = int(input())
    XYP = [tuple(map(int, input().split())) for _ in range(N)]
    X = [x for x, _, _ in XYP]
    Y = [y for _, y, _ in XYP]
    P = [p for _, _, p in XYP]
    INF = 10**18

    def calcCost(AP, A):
        ret = [[INF] * N for _ in range(1 << N)]
        for state in range(1 << N):
            useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
            for i, (a, p) in enumerate(AP):
                cost = abs(a)
                for z in useA:
                    cost = min(cost, abs(a - z))
                ret[state][i] = cost * p
        return ret

    costX = calcCost([(x, p) for x, _, p in XYP], X)
    costY = calcCost([(y, p) for _, y, p in XYP], Y)

    ans = [INF] * (N + 1)
    for state in product((0, 1, 2), repeat=N):
        stateX = 0
        stateY = 0
        M = N
        for s in state:
            stateX <<= 1
            stateY <<= 1
            if s == 0: M -= 1
            if s == 1: stateX |= 1
            if s == 2: stateY |= 1

        cx, cy = costX[stateX], costY[stateY]
        cost, an = 0, ans[M]
        for s in map(min, zip(cx, cy)):
            cost += s
            if cost >= an:
                break
        else:
            ans[M] = cost

    print(*ans, sep='\n')

sol()

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [INF] * (N + 1)
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    cx, cy = costX[stateX], costY[stateY]
    cost, an = 0, ans[M]
    for s in map(min, zip(cx, cy)):
        cost += s
        if cost >= an:
            break
    else:
        ans[M] = cost

print(*ans, sep='\n')

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [INF] * (N + 1)
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    cx, cy = costX[stateX], costY[stateY]
    cost = 0
    for s in map(min, zip(cx, cy)):
        cost += s
        if cost >= ans[M]:
            break
    else:
        ans[M] = cost

print(*ans, sep='\n')

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [[] for _ in range(N + 1)]
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    ans[M].append((costX[stateX], costY[stateY]))

for xy in ans:
    res = INF
    for a in xy:
        res = min(res, sum(map(min, zip(*a))))
    print(res)

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [INF] * (N + 1)
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    cost = sum(map(min, zip(costX[stateX], costY[stateY])))
    if ans[M] > cost:
        ans[M] = cost

print(*ans, sep='\n')

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, p) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost * p
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [INF] * (N + 1)
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    cost = 0
    for i in range(N):
        cost += min(costX[stateX][i], costY[stateY][i])

    if ans[M] > cost:
        ans[M] = cost

print(*ans, sep='\n')

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq, math
from itertools import zip_longest, permutations, combinations, combinations_with_replacement
from itertools import accumulate, dropwhile, takewhile, groupby
from functools import lru_cache
from copy import deepcopy

N = int(input())
X, Y, P = [[0] * N for _ in range(3)]

for i in range(N):
    X[i], Y[i], P[i] = map(int, input().split())

V = [[0] * N for _ in range(1 << N)]
H = [[0] * N for _ in range(1 << N)]

for i in range(1 << N):
    vs, hs = 0, 0
    for j in range(N):
        v = abs(X[j] * P[j])
        h = abs(Y[j] * P[j])
        for k in range(N):
            if ((i >> k) & 1) == 1:
                v = min(v, abs(X[k] - X[j]) * P[j])
                h = min(h, abs(Y[k] - Y[j]) * P[j])
        V[i][j] = v
        H[i][j] = h


def calc(X, Y):

    ret = 0
    for i in range(N):
        ret += min(V[X][i], H[Y][i])

    return ret


ans = [1 << 60] * (N + 1)


def solve(idx, X, Y):
    if idx == N:
        pc = bin(X).count("1") + bin(Y).count("1")
        ans[pc] = min(ans[pc], calc(X, Y))
        return

    solve(idx + 1, X, Y)
    solve(idx + 1, X | (1 << idx), Y)
    solve(idx + 1, X, Y | (1 << idx))


solve(0, 0, 0)

for i in range(N + 1):
    print(ans[i])

Submission status: Time Limit Exceeded

Solution code:
from itertools import product

N = int(input())
XYP = [tuple(map(int, input().split())) for _ in range(N)]
X = [x for x, _, _ in XYP]
Y = [y for _, y, _ in XYP]
P = [p for _, _, p in XYP]
INF = 10**18

def calcCost(AP, A):
    ret = [[INF] * N for _ in range(1 << N)]
    for state in range(1 << N):
        useA = [a for i, a in enumerate(A) if (state & (1 << i)) != 0]
        for i, (a, _) in enumerate(AP):
            cost = abs(a)
            for z in useA:
                cost = min(cost, abs(a - z))
            ret[state][i] = cost
    return ret

costX = calcCost([(x, p) for x, _, p in XYP], X)
costY = calcCost([(y, p) for _, y, p in XYP], Y)

ans = [INF] * (N + 1)
for state in product((0, 1, 2), repeat=N):
    stateX = 0
    stateY = 0
    M = N
    for s in state:
        stateX <<= 1
        stateY <<= 1
        if s == 0: M -= 1
        if s == 1: stateX |= 1
        if s == 2: stateY |= 1

    cx = costX[stateX]
    cy = costY[stateY]
    cost = 0

    for i, p in enumerate(P):
        cost += p * min(cx[i], cy[i])

    if ans[M] > cost:
        ans[M] = cost

print(*ans, sep='\n')