puzzle（1311）《操作》点亮所有的灯

#include<iostream>
using namespace std;

int main()
{	
	int needlighten[10];
	cout << "输入维度" << endl;
	int n;//n阶点亮所有的灯
	cin >> n;
	cout << "点亮第1行，第2行。。。第" << n - 1;
	cout << "行\n最后一行中，从左到右输入状态，1表示黑，0表示亮" << endl;
	for (int nl = 0; nl<n; nl++) cin >> needlighten[nl];
	int biao[12][12];
	for (int number = 1; number<(1<<n); number++)
	{
		int num = number;
		for (int k = 0; k<n; k++)
		{
			biao[1][k + 1] = num % 2;
			num = num / 2;
		}
		for (int i = 0; i<n + 2; i++)for (int j = 0; j<n + 2; j++)
		{
			if (i == 0 || j == 0 || j == n + 1)biao[i][j] = 0;
			else if (i != 1)biao[i][j] = (biao[i - 2][j] + biao[i - 1][j - 1] + biao[i - 1][j] + biao[i - 1][j + 1]) % 2;
		}
		bool buer = 1;
		for (int u = 0; u<n; u++)if (biao[n + 1][u + 1] != needlighten[u])buer = 0;
		if (buer)
		{
			cout << "点第一行的第";
			for (int v = 0; v < n; v++)if (biao[1][v + 1])cout << v + 1 << "  ";
			cout << "个，再逐行完成即可";
		}
		if (buer)break;
	}
	system("pause > nul");
	return 0;
}

有了这个代码，这个游戏玩起来就十分简单了。

但是，这不代表这个游戏就没有研究的价值了，相反，还有非常有意思的事情是代码显示不了的。

POJ 3279 Fliptile

题目：

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

Sample Output

这个题目的意思是输入m，n，输入m*n的0-1矩阵，要求输出1个m*n的0-1矩阵。

除了尺寸不是正方形之外，其他的和我的描述都一样，输入全局状态（0-1矩阵），输出全局操作（0-1矩阵）

题目的意思是，如果有多个可行的方案的话，输出字典序最小的方案。

但是标程里面的字典序有问题，详情：点击打开链接

因为前面是针对我自己玩游戏的情况，而这里是针对ACM编程，所以，虽然是同样的问题，但是代码略有区别。

在这里，我用到了状态压缩，即用1个整数表示1行。

不过这样写出来的代码确实很难看懂，

f(k)表示的是，一个k代表的行操作对应的行增量，即对某一行进行k代表的行操作，那么这一行会有什么影响。

比如说，如果k代表的是a b c d e，那么对应的影响就是a+b a+b+c b+c+d c+d+e d+e

这个影响，可以看成3个影响的叠加：（1）a b c d e（2）b c d e 0（3）0 a b c d

状态压缩之后就是f里面的计算方法了。

代码：

#include<iostream>
using namespace std;

int m, n, a;
int l[17];
int temp[17];

int f(int k)
{
	int t = (1 << n) - 1;
	return (k * 2 ^ k^k / 2)&t;
}

bool ok(int k)
{
	for (int i = 0; i < 17; i++)temp[i] = l[i];
	temp[0] = k;
	for (int i = 1; i < m; i++)
	{
		temp[i] = (temp[i] ^ f(temp[i - 1]));
		temp[i + 1] = (temp[i + 1] ^ temp[i - 1]);
	}
	return temp[m]==f(temp[m-1]);
}

void out(int k)
{
	for (int i = 0; i < 17; i++)temp[i] = l[i];
	temp[0] = k;
	for (int i = 1; i <= m; i++)
	{
		for (int j = 0; j < n; j++)
		{
			cout << (temp[i - 1] >> j) % 2;
			if (j < n - 1)cout << " ";
		}
		cout << endl;

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