CCF Python題解(60分)201604-4 遊戲

# 最短路徑 廣義優先搜尋 BFS
# 所以有時候需要在一些地方踱步，等過了這段時間再前行，就不能簡單地限制為進入過的格不能再進入。
from collections import deque

visited = [[[0 for k in range(301)] for i in range(101)] for j in range(101)]

n, m, t = map(int, input().split())
direct = [[-1, 0], [1, 0], [0, -1], [0, 1]]
for i in range(t)
 
:
    r, c, a, b = map(int, input().split())
    for j in range(a, b + 1):
        visited[r][c][j] = 1


def bfs(x, y):
    q = deque()
    q.append([1, 1, 0])

    while len(q) != 0:
        front = q.popleft()
        if front[0] == x and front[1] == y:
            return front[2]
        for k in range
 
(4):

            row = front[0] + direct[k][0]
            col = front[1] + direct[k][1]
            level = front[2] + 1
            if row < 1 or row > n or col < 1 or col > m or visited[row][col][level]:
                continue
            visited[row][col][level] = 1
            q.append(
 
[row, col, level])


print(bfs(n, m))