题目描述

blablabla

样例

blablabla

算法1

blablabla

时间复杂度

参考文献

Python3 代码

import collections
from sys import stdin
n, m = [int(x) for x in stdin.readline().strip().split(' ')]
done = [[False for _ in range(m)] for _ in range(n)]

migong = []

for i in range(n):
    migong.append( [int(x) for x in stdin.readline().strip().split(' ')] )

def bfs(migong):
    res = 0
    que = collections.deque()
    que.append((0,0,0))
    done[0][0] = True

    dirt = [(1,0),(0,1),(-1,0),(0,-1)]
    while que:
        #size = len(que)
        #for _ in range(size):
        x,y,dis = que.popleft()
        if x == n - 1 and y == m - 1:
            print(dis)
            break
        else:
            for i,j in dirt:
                if 0 <= x+i < n and 0 <= y+j < m and migong[x+i][y+j] == 0 and not done[x+i][y+j]:
                    que.append((x+i,y+j,dis+1))
                    done[x+i][y+j] = True


if __name__ == "__main__":
    bfs(migong)

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla