1. 给定数组nums,执行k次操作，每次选择一个数+1，求k次之后可能最小的全组最大数。
n, k = map(int,input().split()) nums = list(map(int, input().split())) maxn = max(nums) total = sum(nums)+k if total%n == 0: print(max(total//n,maxn)) else: print(max(total//n+1, maxn))
2.给定grid，'0'表示可通过，'1'表示不可通过，在[x,y]处有军队（多处），求所有军队集合的最短时间，不能集合返回-1。
n, m = map(int, input().split())
grid = []
for i in range(n):
row = list(input().strip(''))
grid.append(row)
k = int(input())
cord = []
for j in range(k):
x,y = map(int,input().split())
cord.append([x-1,y-1])
dirs = [[0,-1],[0,1],[-1,0],[1,0]]
def neibors(i,j,visited,grid):
res = []
for dx,dy in dirs:
x = i+ dx
y = j+ dy
if x in range(n) and y in range(m) and not visited[x][y] and grid[x][y] != '1':
res.append([x,y])
return res
def dfs(grid, x, y, cur, visited, new):
new[x][y] = cur
visited[x][y] = True
neibor = neibors(x,y,visited,grid)
if len(neibor) > 0:
for i,j in neibor:
dfs(grid, i, j, cur+1, visited, new)
return new
def update(m1,m2):
for i in range(n):
for j in range(m):
m1[i][j] = max(m1[i][j], m2[i][j])
return m1
result = [[0]*m for _ in range(n)]
for x,y in cord:
new = [[10**10]*m for _ in range(n)]
visited = [[False]*m for _ in range(n)]
this = dfs(grid, x, y, 0, visited, new)
result = update(result, this)
curmin = 10**10
for i in range(n):
for j in range(m):
curmin = min(curmin, result[i][j])
if curmin != 10**10:
print(curmin)
else:
print(-1)
测试例子
3 3 000 101 000 4 1 1 1 3 3 1 3 3
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