AtCoder Beginner Contest 203（Sponsored by Panasonic）D.Pond

题目链接

Problem Statement

The land of a park AtCoder is an N×NN×NN×N grid with east-west rows and north-south columns. The height of the square at the i-th row from the north and j-th column from the west is given as Ai,jA_{i,j}Ai,j.
Takahashi, the manager, has decided to build a square pond occupying K×KK×KK×K squares in this park.To do this, he wants to choose a square section of
K×K squares completely within the park whose median of the heights of the squares is the lowest. Find the median of the heights of the squares in such a section.
Here, the median of the heights of the squares in a K×K section is the height of the (⌊K22⌋+1)(\lfloor \frac{K^2}{2}⌋+1)(⌊2K2⌋+1)-th highest square among the K2K^2K2 squares in the section, where ⌊x⌋⌊x⌋⌊x⌋ is the greatest integer not exceeding xxx.

Sample Input 1

Sample Output 1

Sample Input 2

Sample Output 2

二维前缀和+差分~
做个小转换，题目要求第 ⌊K22⌋+1\lfloor \frac{K^2}{2}⌋+1⌊2K2⌋+1 大，实际上就是求第 ⌈K22⌉\lceil\frac{K^2}{2}\rceil⌈2K2⌉ 小~
如果暴力的话时间复杂度 O(N∗N∗K∗K)O(N*N*K*K)O(N∗N∗K∗K)，显然不行，比赛中我还想了用数据结构存，但是复杂度也有 O(N∗N∗K)O(N*N*K)O(N∗N∗K)，还是会爆~
回到一开始，找一个 K∗KK*KK∗K 的矩阵第 ⌈K22⌉\lceil\frac{K^2}{2}\rceil⌈2K2⌉ 小，即这个矩阵中有 ⌈K22⌉\lceil\frac{K^2}{2}\rceil⌈2K2⌉ 个元素小于等于这个值，那么我们很容易想到二分，那怎么快速判断某个 K∗KK*KK∗K 矩阵中小于等于某个数的个数呢，离散化+二维前缀和即可，对二分的值 midmidmid，首先离散化，当 Ai,j≤midA_{i,j}\leq midAi,j≤mid，则置为 111，反之置为 000，然后通过二维前缀和可以快速计算出一个 K∗KK*KK∗K 矩阵中大于 midmidmid 的元素个数，暴力判断即可，复杂度 O(N∗N∗log⁡1e9)O(N*N*\log{1e9})O(N∗N∗log1e9)，AC代码如下：

n, k = map(int, input().split())
a = [[] for _ in range(805)]
b = [[0 for _ in range(805)] for _ in range(805)]
c = [[0 for _ in range(805)] for _ in range(805)]
for i in range(1, n + 1):a[i] = list(map(int, input().split()))# print(a[i])
l, r = 0, 10 ** 9
while l <= r:mid = l + r >> 1for i in range(1, n + 1):for j in range(1, n + 1):b[i][j] = 0 if a[i][j - 1] > mid else 1for i in range(1, n + 1):for j in range(1, n + 1):c[i][j] = c[i - 1][j] + c[i][j - 1] - c[i - 1][j - 1] + b[i][j]if c[n][n] < (k * k - 1) // 2 + 1:l = mid + 1continueflag = 0for i in range(k, n + 1):for j in range(k, n + 1):if c[i][j] - c[i - k][j] - c[i][j - k] + c[i - k][j - k] >= (k * k - 1) // 2 + 1:flag = 1breakif flag:breakif flag:r = mid - 1else:l = mid + 1
print(l)