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POJ - 3279 Fliptile 【状态压缩、dfs】
题目简述 Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored翻转问题开关问题 poj 3279 fliptile
题目： https://vjudge.net/problem/POJ-3279 参考的题解： POJ-3279 经典翻转问题_越努力越幸运—liupu-CSDN博客 翻转吧！POJ 3279! - 知乎 (zhihu.com) 思路：把第一排的01串给枚举出来（直接按照字典序来） 从第二排开始把每种情况用b存下来 检查每种情况最后一排有没有黑色 如果全白从Fliptile POJ - 3279
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M × N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored blacFliptile
题意：给一个矩阵，矩阵内元素非0即1且各元素可以翻转，0可以翻转为1，1可以翻转为0，翻转某一个元素时，它的上下左右个元素也会翻转 （注：受影响翻转的不会继续影响其它的），先问是否可以成功将元素全部置0，若可以，输出翻转次数最小的，若有多个答案，按字典序输出 思路：首先因为翻转一个元素会造成许多POJ-3279 Fliptile
D - Fliptile (POJ - 3279) using namespace std; typedef long long ll; const int inf = 0x3f3f3f3f; const int N = 15+5; int b[N][N] = {{0}}, c[N][N] = {{0}}, a[N][N] = {{0}}; int n, m, ans = inf; int dx[]= { 1, -1, 0, 0, 0}; int dy[]= { 0, 0, 1, -1Fliptile （二进制压缩）
题目链接：http://poj.org/problem?id=3279 题目大意：有一个n*m的棋盘，0表示白色，1表示黑色。每次可以翻转当前位置，它的上下左右四个位置也会被相应翻转。问最少翻转多少次会使所有棋面显示为白色，并给出需要翻转的位置，0表示不翻转，1表示翻转。 思路：第一行的翻转状态决定了你第二行Fliptile (dfs+二进制压缩)
Farmer John knows that an intellectually satisfied cow is a happy cow who will give more milk. He has arranged a brainy activity for cows in which they manipulate an M× N grid (1 ≤ M ≤ 15; 1 ≤ N ≤ 15) of square tiles, each of which is colored bl