题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

python 代码

import sys
import collections
from copy import deepcopy
n = int(input().split()[0])
#print(n)


moves = [(0,0),(0,1),(1,0),(0,-1),(-1,0)]
def turn(nums, i,j):
    for x,y in moves:
        newx, newy = i + x, j  + y
        if 0 <= newx < 5 and 0<= newy < 5:
            nums[newx][newy] ^= 1

def cal(lights, k):
    res = 0 
    for i in range(5):  
        if 1 == k >> i & 1:  # 枚举状态  k 右移i位和1& 这句话的作用就是判断当前位是否是1  1表示当前位要操作 0 表示不操作 这个操作的含义就是点击， 要区别亮灭的含义
            turn(lights, 0, i)
            res += 1  # ok确定好第一行了

    for i in range(1, 5):  # 依据第一行来,从第二行开始
        for j in range(5):
            if lights[i-1][j] == 0:
                turn(lights, i, j)
                res += 1
    for j in range(5):
        if lights[-1][j] == 0: return 7
    return res

'''
nums = []    
for _ in range(6):  # read data
    tmp = []
    arr = sys.stdin.readline().split()
    print(arr)
    if not arr: continue
    else:
        arr = arr[0]
    for j in range(5):
        tmp.append(int(arr[j]))
    nums.append(tmp)
print(nums)
'''   



while n > 0:
    res = float("inf")
    nums = []
    for _ in range(5):  # read data
        tmp = []
        arr = sys.stdin.readline().split()
        if not arr: continue
        else:
            arr = arr[0]
        for j in range(5):
            tmp.append(int(arr[j]))
        nums.append(tmp)
    for k in range(1 << 5):  # 枚举第一行按还是不按的状态  
        count = cal(deepcopy(nums), k)
        res = min(res, count)
    a = sys.stdin.readline().split()

    if res <=6 : print(res)
    else:
        print(-1)
    n -=1