【2020秋招笔试】京东-消消乐

最新推荐文章于 2024-05-29 15:44:39 发布

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本文深入探讨了一款消消乐游戏的实现算法，详细解释了如何通过深度优先搜索（DFS）来判断并消除相邻的相同元素，以及如何计算游戏结束后剩余的不可消除元素数量。文章提供了完整的Python代码示例，包括网格状态更新、岛屿面积计算和消除操作等关键步骤。

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消消乐，>= 3个可消除掉，重力掉落，问最后最少剩几个

样例：（输入必须5 * 5，只有1，2，3三种类型）
3 1 2 1 1
1 1 1 1 2
1 1 1 1 1
1 1 1 1 1
3 1 2 2 2
输出：4
因为，消除最后结果为：
3 0 2 0 0
0 0 0 0 2
3 0 0 0 0

import sys
from collections import defaultdict
direction = [[1,0],[0,1],[-1,0],[0,-1]]
grid = []
for i in range(5):
    line = sys.stdin.readline().strip()
    nums = list(map(int, line.split()))
    grid.append(nums)

def del_grid():
    global grid
    s = []
    for i in range(len(grid)):
        if(grid[i].count(0) != 5):
            s.append(grid[i])        
    return s

def judge(r, c):
    global grid     
    if(r < 0 or r >= len(grid) or c < 0 or c >= len(grid[0])):
        return False
    return True

def maxAreaOfIsland(): 
    global grid        
    Area = 0
    i, j = 0, 0    
    while(i < len(grid) and j < len(grid[0])):
        vis = [[0 for k in range(len(grid[0]))] for w in range(len(grid))]
        if(grid[i][j] != 0):
            tar = grid[i][j]            
            Area = dfs(i, j, tar, vis)            
            if(Area >= 3):
                a = dfs_(i, j, tar)
                grid = del_grid()
                maxAreaOfIsland()
        j = j + 1
        if(j == len(grid[0])):
            i += 1
            j = 0
    return

def dfs_(r, c, tar) :
    global grid
    if (judge(r,c) == False or grid[r][c] != tar):
        return 0
    grid[r][c] = 0
    area = 1
    for d in direction:
        area += dfs_(r + d[0], c + d[1], tar)
    return area

def dfs(r, c, tar, vis):
    global grid        
    if (judge(r,c) == False or grid[r][c] != tar or vis[r][c] == 1):
        return 0    
    vis[r][c] = 1
    area = 1
    for d in direction:
        area += dfs(r + d[0], c + d[1], tar, vis)
    return area

t = 0
maxAreaOfIsland()
for i in range(len(grid)):
    t += (grid[i].count(1) + grid[i].count(2)+ grid[i].count(3))
print(t)