[BaekJoon Online Judge] 12100 – 2048 (Easy)

문제

https://www.acmicpc.net/problem/12100

소스코드

문제에서 최대 다섯 번의 방향 전환이 가능하다고 제약을 걸었기 때문에, 5중 루프를 돌려서 가능한 모든 경우의 수를 만들었다.

앞서 만든 경우에 수에 대해, 문제에서 설명한 2048 게임룰(방향 전환에 따라 블록이 합쳐지는 조건)을 노가다로 작성해서 문제를 해결할 수 있었다.

from copy import deepcopy
def game(direction):
    global b
    if direction == 0: # 오른쪽
        for i in range(N):
            tmp = []
            for j in range(N):
                if b[i][j] != 0:
                    tmp.append(b[i][j])
            for _ in range(N-len(tmp)):
                tmp.insert(0,0)
            b[i] = tmp
        for i in range(N):
            for j in range(N-1,0,-1):
                if b[i][j] == b[i][j-1]:
                    b[i][j] *= 2
                    for k in range(j-1,0,-1):
                        tmp = b[i][k]
                        b[i][k] = b[i][k-1]
                        b[i][k-1] = tmp
                    b[i][0] = 0 
    if direction == 1: # 아랫쪽
        for i in range(N):
            tmp = []
            for j in range(N):
                if b[j][i] != 0:
                    tmp.append(b[j][i])
            for j in range(N-len(tmp)):
                b[j][i] = 0
            for j in range(N-len(tmp),N):
                b[j][i] = tmp.pop(0)
        for i in range(N):
            for j in range(N-1,0,-1):
                if b[j][i] == b[j-1][i]:
                    b[j][i] *= 2
                    for k in range(j-1,0,-1):
                        tmp = b[k][i]
                        b[k][i] = b[k-1][i]
                        b[k-1][i] = tmp
                    b[0][i] = 0 
    if direction == 2: # 왼쪽
        for i in range(N):
            tmp = []
            for j in range(N):
                if b[i][j] != 0:
                    tmp.append(b[i][j])
            for _ in range(N-len(tmp)):
                tmp.append(0)
            b[i] = tmp
        for i in range(N):
            for j in range(N-1):
                if b[i][j] == b[i][j+1]:
                    b[i][j] *= 2
                    for k in range(j+1,N-1):
                        tmp = b[i][k]
                        b[i][k] = b[i][k+1]
                        b[i][k+1] = tmp
                    b[i][N-1] = 0 
    if direction == 3: # 윗쪽
        for i in range(N):
            tmp = []
            for j in range(N):
                if b[j][i] != 0:
                    tmp.append(b[j][i])
            k = len(tmp)
            for j in range(k):
                b[j][i] = tmp.pop(0)
            for j in range(k,N):
                b[j][i] = 0
        for i in range(N):
            for j in range(N-1):
                if b[j][i] == b[j+1][i]:
                    b[j][i] *= 2
                    for k in range(j+1,N-1):
                        tmp = b[k][i]
                        b[k][i] = b[k+1][i]
                        b[k+1][i] = tmp
                    b[N-1][i] = 0 

N = int(input())
block = [list(map(int,input().split())) for _ in range(N)]

case = []
for i in range(4):
    for j in range(4):
        for k in range(4):
            for l in range(4):
                for m in range(4):
                    case.append([i,j,k,l,m])
ans = 0
for c in case:
    b = deepcopy(block)
    for direction in c: game(direction)
    tmp = 0
    for i in range(N):
        for j in range(N):
            if b[i][j] > tmp:
                tmp = b[i][j]
    if tmp > ans:
        ans = tmp
print(ans)