poj1222 EXTENDED LIGHTS OUT 高斯消元

最新推荐文章于 2019-12-09 19:29:44 发布

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#高斯消元 #数学

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本文提供了一种求解五行六列灯谜游戏的算法，通过数学建模和高斯消元法，实现了从任意初始状态到全灯熄灭的目标。详细解释了每一步的操作逻辑，并提供了代码实现，旨在帮助读者理解并应用到类似问题中。

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EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6409		Accepted: 4207

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

Source

Greater New York 2002

题意：给定一个5*6由灯组成的方阵，对一个灯进行操作，它及其周围4个灯都会变化，即由暗到明或由明到暗，问如何操作使得灯全变暗。输出一个矩阵，1为操作，0为不操作。

思路：对于每个灯的亮暗有影响的开关只有它及其附近十字形内的5个开关，所以对于每个灯可以列一个方程，设Xi为第i个开关是否有操作，若初始时i灯暗，则Xi ^ Xi-1 ^ Xi+1 ^ Xi-6 ^ Xi+6 = 0, 亮则Xi ^ Xi-1 ^ Xi+1 ^ Xi-6 ^ Xi+6 = 1,即等式右边为初始输入状态。之后进行高斯消元，得出每个Xi,最后按矩阵输出，详见代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=50;
int a[MAXN][MAXN];
void Gauss(int n)
{
    for(int i=0;i<n;i++)
    {
        int r=i;
        for(int j=i;j<n;j++)
            if(a[j][i])
            {
                 r=j;
                 break;
            }

        if(r!=i)
            for(int j=0;j<=n;j++) swap(a[r][j],a[i][j]);
        for(int k=i+1;k<n;k++)
            if(a[k][i])
                for(int j=i;j<=n;j++)
                    a[k][j]^=a[i][j];
    }
    for(int i=n-1;i>=0;i--)
        for(int j=i+1;j<n;j++)
            a[i][n]^=(a[i][j] && a[j][n]);
}
int main()
{
    //freopen("text.txt","r",stdin);
    int T,kase=0;
    scanf("%d",&T);
    while(T--)
    {
        kase++;
        memset(a,0,sizeof(a));
        for(int i=0;i<30;i++)
        {
            scanf("%d",&a[i][30]);
            a[i][i]=1;
            if(i%6!=0)
                a[i][i-1]=1;
            if(i%6!=5)
                a[i][i+1]=1;
            if(i>5)
                a[i][i-6]=1;
            if(i<24)
                a[i][i+6]=1;
        }
        Gauss(30);
        printf("PUZZLE #%d\n",kase);
        for(int i=0;i<30;i++)
        {
            if(i%6==0)
                printf("%d",a[i][30]);
            else
                printf(" %d",a[i][30]);
            if(i%6==5)
                printf("\n");
        }
    }
    return 0;
}