【代码超详解】POJ 3279 Fliptile（开关问题，DFS）

一、题目描述

Fliptile

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 25691 Accepted: 9099

Description

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 black on one side and white on the other side.

As one would guess, when a single white tile is flipped, it changes to black; when a single black tile is flipped, it changes to white. The cows are rewarded when they flip the tiles so that each tile has the white side face up. However, the cows have rather large hooves and when they try to flip a certain tile, they also flip all the adjacent tiles (tiles that share a full edge with the flipped tile). Since the flips are tiring, the cows want to minimize the number of flips they have to make.

Help the cows determine the minimum number of flips required, and the locations to flip to achieve that minimum. If there are multiple ways to achieve the task with the minimum amount of flips, return the one with the least lexicographical ordering in the output when considered as a string. If the task is impossible, print one line with the word “IMPOSSIBLE”.

Input

Line 1: Two space-separated integers: M and N
Lines 2…M+1: Line i+1 describes the colors (left to right) of row i of the grid with N space-separated integers which are 1 for black and 0 for white

Output

Lines 1…M: Each line contains N space-separated integers, each specifying how many times to flip that particular location.

Sample Input

4 4
1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

Sample Output

0 0 0 0
1 0 0 1
1 0 0 1
0 0 0 0

Source

USACO 2007 Open Silver

二、算法分析说明与代码编写指导

本指导是在优快云的某篇博客的基础上补充而来的（原博已找不到）。
题目大意是，让奶牛翻转（flip） M × N 的砖块矩阵，使得每一个砖块都变成白色。砖块有两面，分别是黑（1）白（0）两色。因为牛蹄太大，翻转一个砖块会令上下左右四个砖块一起被翻转。
要求输出翻转次数最小的方案。输出的方案中，1 代表翻转，0 代表不翻转。
由于 M、N ≤ 15，数据量足够小，可以进行 DFS 暴力枚举。
首先我们仅需枚举第一行的翻转情况，剩余行是否需要翻转可以递推出来：
AC 代码的第 32、33、34、35 行

	I = 1 << n;
	for (int i = 0; i < I; ++i) {
   
   
		for (int j = 1; j <= m; ++j)fill(f[j] + 1, f[j] + n + 1, 0);
		for (int j = 0; j < n; ++j)f[1][n - j] = i >> j & 1;