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3240. 最少翻转次数使二进制矩阵回文 II
题意
给你一个 m x n 的二进制矩阵 grid 。
如果矩阵中一行或者一列从前往后与从后往前读是一样的,那么我们称这一行或者这一列是 回文 的。
你可以将 grid 中任意格子的值 翻转 ,也就是将格子里的值从 0 变成 1 ,或者从 1 变成 0 。
请你返回 最少 翻转次数,使得矩阵中 所有 行和列都是 回文的 ,且矩阵中 1 的数目可以被 4 整除 。
思路
并查集+思维+dp
- 如果一个数变的话,为了保持回文,那么行对称和列对称的两个数也得变(其实是四个数),所以这就形成了一个集合,且集合与集合之间互不影响
- 即使用并查集维护,并维护集合中1的个数和集合大小
- 使用dp状态机,dp[i]表示:
1的数量 %4=i 的最小操作次数
,当遇到根节点,对集合针对0和1都进行翻转,因为集合只能全为0和1 - 拓展性更强,把4的倍数改成 2 3 4 5 6 7 8 9 的倍数都没关系
代码
python
'''
Author: NEFU AB-IN
Date: 2024-08-07 10:04:41
FilePath: \LeetCode\3240\3240.py
LastEditTime: 2024-08-07 11:06:07
'''
# 3.8.19 import
import random
from collections import Counter, defaultdict, deque
from datetime import datetime, timedelta
from functools import lru_cache, reduce
from heapq import heapify, heappop, heappush, nlargest, nsmallest
from itertools import combinations, compress, permutations, starmap, tee
from math import ceil, comb, fabs, floor, gcd, hypot, log, perm, sqrt
from string import ascii_lowercase, ascii_uppercase
from sys import exit, setrecursionlimit, stdin
from typing import Any, Callable, Dict, List, Optional, Tuple, TypeVar, Union
# Constants
TYPE = TypeVar('TYPE')
N = int(2e5 + 10)
M = int(20)
INF = int(1e12)
OFFSET = int(100)
MOD = int(1e9 + 7)
# Set recursion limit
setrecursionlimit(int(2e9))
class Arr:
array = staticmethod(lambda x=0, size=N: [x() if callable(x) else x for _ in range(size)])
array2d = staticmethod(lambda x=0, rows=N, cols=M: [Arr.array(x, cols) for _ in range(rows)])
graph = staticmethod(lambda size=N: [[] for _ in range(size)])
class Math:
max = staticmethod(lambda a, b: a if a > b else b)
min = staticmethod(lambda a, b: a if a < b else b)
class IO:
input = staticmethod(lambda: stdin.readline().rstrip("\r\n"))
read = staticmethod(lambda: map(int, IO.input().split()))
read_list = staticmethod(lambda: list(IO.read()))
class Std:
class UnionFind:
"""Union-Find data structure."""
def __init__(self, n):
self.n = n
self.parent = list(range(n)) # Parent pointers
self.rank = Arr.array(1, n) # Rank (approximate tree height)
self.size = Arr.array(1, n) # Size arrays for each node
self.cnt_1 = Arr.array(0, n)
def find(self, p):
"""Find the root of the element p with path compression."""
if self.parent[p] != p:
self.parent[p] = self.find(self.parent[p]) # Path compression
return self.parent[p]
def union(self, p, q):
"""Union the sets containing p and q using union by rank and merge data if available."""
rootP = self.find(p)
rootQ = self.find(q)
if rootP != rootQ:
# Union by rank
if self.rank[rootP] > self.rank[rootQ]:
self.parent[rootQ] = rootP
self.size[rootP] += self.size[rootQ]
self.cnt_1[rootP] += self.cnt_1[rootQ]
return rootP
elif self.rank[rootP] < self.rank[rootQ]:
self.parent[rootP] = rootQ
self.size[rootQ] += self.size[rootP]
self.cnt_1[rootQ] += self.cnt_1[rootP]
return rootQ
else:
self.parent[rootQ] = rootP
self.size[rootP] += self.size[rootQ]
self.cnt_1[rootP] += self.cnt_1[rootQ]
self.rank[rootP] += 1
return rootP
return rootP # They are already in the same set
# --------------------------------------------------------------- Division line ------------------------------------------------------------------
class Solution:
def minFlips(self, grid: List[List[int]]) -> int:
m, n = len(grid), len(grid[0])
uf = Std.UnionFind(m * n)
def index(i, j): return i * n + j
for i in range(m):
for j in range(n):
uf.cnt_1[index(i, j)] = grid[i][j]
for i in range(m):
for j in range(n):
uf.union(index(i, j), index(i, n - j - 1))
uf.union(index(i, j), index(m - i - 1, j))
dp = [0, INF, INF, INF] # dp[i]表示:1的数量%4=i的最小操作次数
for i in range(m):
for j in range(n):
if uf.find(index(i, j)) == index(i, j):
f = [INF] * 4
for x in range(4): # 每个连通块内,要么全1,要么全0
# 全变为0,则需要将该连通块的所有1翻转
f[(x + 0) % 4] = Math.min(f[(x + 0) % 4], dp[x] + uf.cnt_1[index(i, j)])
# 全变为1,则需要将该连通块的所有0翻转,0的数量是总数量减1的数量
f[(x + uf.size[index(i, j)]) % 4] = Math.min(f[(x + uf.size[index(i, j)]) % 4], dp[x] + uf.size[index(i, j)] - uf.cnt_1[index(i, j)])
dp = f
return dp[0]