题目链接：http://poj.org/problem?id=1222
Description
In an extended version of the game Lights Out, is a puzzle with 5 rows of 6 buttons each (the actual puzzle has 5 rows of 5 buttons each). Each button has a light. When a button is pressed, that button and each of its (up to four) neighbors above, below, right and left, has the state of its light reversed. (If on, the light is turned off; if off, the light is turned on.) Buttons in the corners change the state of 3 buttons; buttons on an edge change the state of 4 buttons and other buttons change the state of 5. For example, if the buttons marked X on the left below were to be pressed,the display would change to the image on the right.
The aim of the game is, starting from any initial set of lights on in the display, to press buttons to get the display to a state where all lights are off. When adjacent buttons are pressed, the action of one button can undo the effect of another. For instance, in the display below, pressing buttons marked X in the left display results in the right display.Note that the buttons in row 2 column 3 and row 2 column 5 both change the state of the button in row 2 column 4,so that, in the end, its state is unchanged.
Note:
1. It does not matter what order the buttons are pressed.
2. If a button is pressed a second time, it exactly cancels the effect of the first press, so no button ever need be pressed more than once.
3. As illustrated in the second diagram, all the lights in the first row may be turned off, by pressing the corresponding buttons in the second row. By repeating this process in each row, all the lights in the first
four rows may be turned out. Similarly, by pressing buttons in columns 2, 3 ?, all lights in the first 5 columns may be turned off.
Write a program to solve the puzzle.
Input
The first line of the input is a positive integer n which is the number of puzzles that follow. Each puzzle will be five lines, each of which has six 0 or 1 separated by one or more spaces. A 0 indicates that the light is off, while a 1 indicates that the light is on initially.
Output
For each puzzle, the output consists of a line with the string: "PUZZLE #m", where m is the index of the puzzle in the input file. Following that line, is a puzzle-like display (in the same format as the input) . In this case, 1's indicate buttons that must be pressed to solve the puzzle, while 0 indicate buttons, which are not pressed. There should be exactly one space between each 0 or 1 in the output puzzle-like display.
Sample Input
2 0 1 1 0 1 0 1 0 0 1 1 1 0 0 1 0 0 1 1 0 0 1 0 1 0 1 1 1 0 0 0 0 1 0 1 0 1 0 1 0 1 1 0 0 1 0 1 1 1 0 1 1 0 0 0 1 0 1 0 0
Sample Output
PUZZLE #1 1 0 1 0 0 1 1 1 0 1 0 1 0 0 1 0 1 1 1 0 0 1 0 0 0 1 0 0 0 0 PUZZLE #2 1 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 0 0 1 1 0 1 0 1 1 0 1 1 0 1
题意：
有T组 5*6的灯阵，用1表示灯亮，0表示灯灭。每个灯的位置有一个开关，刚开始开关全是关的，用0表示开关关，用1表示开关开。每次切换开关的状态（即从1->0或从0->1 ) 会改变这个开关上下左右四个位置上灯的状态，灯亮则灭，灯灭则亮（如果有）。要求给让每组 所有灯都熄灭时的开关的状态。
注意：
对同一个开关操作第二次会抵消第一次的操作，而进行三次操作相当于只进行一次操作。所以对同一个开关惊醒多次操作是没有意义的。
代码：
#include <iostream>
using namespace std;
char oriLights[5];
char lights[5];
char result[5];
int GetBit(char c,int i)
{
return (c >> i) & 1 ;
}
void SetBit(char &c,int i,int v)
{
if( v)
c |= (1 << i);
else
c &= ~(1 << i);
}
void FlipBit(char &c,int i)
{
c ^= (1 << i);
}
void OutputResult(int t,char result[])
{
cout << "PUZZLE #" << t << endl;
for(int i=0;i<5;i++)
{
for(int j=0;j<6;j++)
{
cout << GetBit(result[i],j);
if(j < 5)
cout << " ";
}
cout << endl;
}
}
int main()
{
int T;
cin >> T;
for(int t=1;t<=T;t++)
{
for(int i=0;i<5;i++)
for(int j=0;j<6;j++){
int s;
cin >> s;
SetBit(oriLights[i],j,s);
}
for(int n=0;n<64;n++) // 每一行开关有64种可能，对应十进制数为 0~63
{
int switchs = n;
memcpy(lights,oriLights,sizeof(oriLights)); //
for(int i=0;i<5;i++)
{
result[i] = switchs;
for(int j=0;j<6;j++)
{
if(GetBit(switchs,j))
{
if(j > 0)
FlipBit(lights[i],j-1);
FlipBit(lights[i],j);
if(j < 5)
FlipBit(lights[i],j+1);
}
}
if(i < 4)
lights[i+1] ^= switchs; // 位
switchs = lights[i]; // *** 为了让第i行亮着的灯熄灭，即在确定第i+1行的灯状态后，第i行的灯已全部熄灭
}
if(lights[4] == 0)
{
OutputResult(t,result);
break;
}
}
}
return 0;
}