目录
一、3194. 最小元素和最大元素的最小平均值
二、3195. 包含所有 1 的最小矩形面积 I
三、3196. 最大化子数组的总成本
四、3197. 包含所有 1 的最小矩形面积 II
博主在比赛时只过了前两题。剩下跟着灵神做，来自视频：
【状态机 DP【力扣周赛 403】】 https://www.bilibili.com/video/BV1MZ421M74P/?share_source=copy_web&vd_source=dc0e55cfae3b304619670a78444fd795
一、3194. 最小元素和最大元素的最小平均值
1.自写：
class Solution:def minimumAverage(self, nums: List[int]) -> float:nums.sort()ans = infl, r = 0, len(nums) - 1while l <= r:ans = min(ans, nums[l] + nums[r])l += 1r -= 1return ans / 22.灵神：
class Solution:def minimumAverage(self, nums: List[int]) -> float:n = len(nums)nums.sort()ans = inffor i in range(n // 2):j = n - 1 - ians = min(ans, nums[i] + nums[j])return ans / 2二、3195. 包含所有 1 的最小矩形面积 I
1.自写：
我每次一遇到这种题，就只一股脑用前缀和（笑哭）。
class Solution:def minimumArea(self, grid: List[List[int]]) -> int:# 前缀和n, m = len(grid), len(grid[0])mx = 0ans = 0flag = (0, 0)add = [[0 for _ in range(m + 1)] for _ in range(n + 1)]for i in range(n):for j in range(m):add[i + 1][j + 1] = add[i][j + 1] + add[i + 1][j] - add[i][j] + grid[i][j]if add[i + 1][j + 1] > mx:mx = add[i + 1][j + 1]ans = (i + 1) * (j + 1)flag = ((i + 1), (j + 1))# 找到前面空的i = 1while i <= n:if add[i][-1] != 0:breaki += 1j = 1while j <= m:if add[-1][j] != 0:breakj += 1ans -= (i - 1) * flag[1] + (j - 1) * flag[0] - (i - 1) * (j - 1)return ans2.灵神：
class Solution:def minimumArea(self, grid: List[List[int]]) -> int:# 边界left, right = len(grid[0]), 0up, down = len(grid), 0for i, row in enumerate(grid):for j, x in enumerate(row):if x:left = min(left, j)right = max(right, j)up = min(up, i)down = i # 从上到下遍历return (down - up + 1) * (right - left + 1)三、3196. 最大化子数组的总成本
当时第一反应是dp，但是我当时想的是前缀，没能解决。
灵神：
附一个自己画的状态转移的简单图解：
（1）递归
class Solution:def maximumTotalCost(self, nums: List[int]) -> int:# dp# 后缀# 递归@cachedef dfs(i: int, j: int) -> int:if i == len(nums):return 0if j == 0:return dfs(i + 1, 1) + nums[i]return max(dfs(i + 1, 1) + nums[i], dfs(i + 1, 0) - nums[i])return dfs(0, 0) # 第一位肯定为正(2)递推，一比一翻译
class Solution:def maximumTotalCost(self, nums: List[int]) -> int:# dp# 后缀# 递推，一比一翻译n = len(nums)f = [[0, 0] for _ in range(n + 1)]for i in range(n - 1, -1, -1):f[i][0] = f[i + 1][1] + nums[i]f[i][1] = max(f[i + 1][1] + nums[i], f[i + 1][0] - nums[i])return f[0][0]（3）递推，滑动窗口
class Solution:def maximumTotalCost(self, nums: List[int]) -> int:# dp# 后缀# 递推，滑动窗口n = len(nums)j0, j1 = 0, 0# 切片nums[::-1]，会消耗额外空间# reversed()生成迭代器，不会消耗额外空间for x in reversed(nums):j0, j1 = j1 + x, max(j1 + x, j0 - x)return j0四、3197. 包含所有 1 的最小矩形面积 II
听了讲解后自己写的代码，旋转90度的代码参考视频。
class Solution:def minimumSum(self, grid: List[List[int]]) -> int:n, m = len(grid), len(grid[0])return min(self.f(grid, n, m), self.f(self.rotate(grid, n, m), m, n))def cover(self, grid: List[List[int]], left: int, right: int, up: int, down: int) -> int:# 计算最小面积# 闭区间# 注意区域内可能没有1，区别第二题# 要么返回时判断，要么初始化时设置大一点l, r = right, leftu, d = down, upfor i in range(up, down + 1):for j in range(left, right + 1):if grid[i][j]:l = min(l, j)r = max(r, j)u = min(u, i)d = ireturn (r - l + 1) * (d - u + 1) if r >= l and d >= u else 0def f(self, grid: List[List[int]], n: int, m: int) -> int:ans = 901# 划分横着三个区域if n >= 3:for line1 in range(n - 2):for line2 in range(line1 + 1, n - 1):m1 = self.cover(grid, 0, m - 1, 0, line1)m2 = self.cover(grid, 0, m - 1, line1 + 1, line2)m3 = self.cover(grid, 0, m - 1, line2 + 1, n - 1)ans = min(ans, m1 + m2 + m3)if n >= 2 and m >= 2:for line1 in range(n - 1):for line2 in range(m - 1):# 划分上横下二区域m1 = self.cover(grid, 0, m - 1, 0, line1)m2 = self.cover(grid, 0, line2, line1 + 1, n - 1)m3 = self.cover(grid, line2 + 1, m - 1, line1 + 1, n - 1)ans = min(ans, m1 + m2 + m3)# 划分下横上二区域m1 = self.cover(grid, 0, line2, 0, line1)m2 = self.cover(grid, line2 + 1, m - 1, 0, line1)m3 = self.cover(grid, 0, m - 1, line1 + 1, n - 1)ans = min(ans, m1 + m2 + m3)return ansdef rotate(self, grid: List[List[int]], n: int, m: int) -> List[List[int]]:# 顺时针旋转90度ans = [[0 for _ in range(n)] for _ in range(m)]for i in range(n):for j in range(m):ans[j][n - 1 - i] = grid[i][j]return ans参考评论区评论（. - 力扣（LeetCode）），事先使用一个覆盖框预处理（想起了R-CNN），修改代码如下：
class Solution:def minimumSum(self, grid: List[List[int]]) -> int:n, m = len(grid), len(grid[0])return min(self.f(grid, n, m), self.f(self.rotate(grid, n, m), m, n))def cover(self, grid: List[List[int]], left: int, right: int, up: int, down: int) -> int:# 计算最小范围# 闭区间l, r = right, leftu, d = down, upfor i in range(up, down + 1):for j in range(left, right + 1):if grid[i][j]:l = min(l, j)r = max(r, j)u = min(u, i)d = ireturn l, r, u, ddef minArea(self, grid: List[List[int]], l: int, r: int, u: int, d: int) -> int:# 注意区域内可能没有1，区别第二题# 要么返回时判断，要么初始化时设置大一点l, r, u, d = self.cover(grid, l, r, u, d) # 找出最小覆盖范围return (r - l + 1) * (d - u + 1) if r >= l and d >= u else 0def f(self, grid: List[List[int]], n: int, m: int) -> int:# 先事先筛选一个大矩形框left, right, up, down = self.cover(grid, 0, m - 1, 0, n - 1)# left -> 0, right -> m - 1, up -> 0, down -> n - 1ans = 901# 划分横着三个区域if n >= 3:for line1 in range(up, down - 1):for line2 in range(line1 + 1, down):m1 = self.minArea(grid, left, right, up, line1)m2 = self.minArea(grid, left, right, line1 + 1, line2)m3 = self.minArea(grid, left, right, line2 + 1, down)ans = min(ans, m1 + m2 + m3)if n >= 2 and m >= 2:for line1 in range(up, down):for line2 in range(left, right):# 划分上横下二区域m1 = self.minArea(grid, left, right, up, line1)m2 = self.minArea(grid, left, line2, line1 + 1, down)m3 = self.minArea(grid, line2 + 1, right, line1 + 1, down)ans = min(ans, m1 + m2 + m3)# 划分下横上二区域m1 = self.minArea(grid, left, line2, up, line1)m2 = self.minArea(grid, line2 + 1, right, up, line1)m3 = self.minArea(grid, left, right, line1 + 1, down)ans = min(ans, m1 + m2 + m3)return ansdef rotate(self, grid: List[List[int]], n: int, m: int) -> List[List[int]]:# 顺时针旋转90度ans = [[0 for _ in range(n)] for _ in range(m)]for i in range(n):for j in range(m):ans[j][n - 1 - i] = grid[i][j]return ans灵神还有一种使用dp进行预处理降低时间复杂度的方法，但是我没学，感觉这暂时不是我能学会的，感兴趣的可以看他的题解（. - 力扣（LeetCode））。
完
感谢你看到这里！一起加油吧！
