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【POJ 1222】EXTENDED LIGHTS OUT

来源：程序员人生发布时间：2015-04-17 08:38:44 阅读次数：3652次

EXTENDED LIGHTS OUT

Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 7333 Accepted: 4792
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

高斯消元解异或方程组。

首先把数组标号为 $id[i][j]$ ，后文所说的第 $k$ 个就是指他的标号 $id[i][j]=k$ 。

①我们用 $a[i][j]$ 表示读入的方格；

② $f[i][j]$ 表示第 $i$ 个方格对第 $j$ 个方格有影响，即你对第 $i$ 个方格履行开关操作时， $j$ 方格也被操作了（ $j$ 在 $i$ 的10字形区域内）；

③最后的答案数组为 $ans[i]$ ， $ans[i]=1$ 表示第 $i$ 个方格被履行操作了；

④ $tot=n*m$ 。

对1个方格 $(i,j)$ ，最后状态为 $0$ ，设 $id[i][j]=k$ ，我们可以列出这样的式子

a [i] [j] + f [k] [1] ? a n s [1] + f [k] [2] ? a n s [2] + . . . f [k] [t o t] ? a n s [t o t] = 0 (m o d 2)

$a[i][j]+f[k][1]*ans[1]+f[k][2]*ans[2]+...f[k][tot]*ans[tot]=0 (mod 2)$

由于对方程 $mod$ $2$ 相当于取异或，我们可以把 $a[i][j]$ 移到右侧，方程变成：

f [k] [1] ? a n s [1] + f [k] [2] ? a n s [2] + . . . f [k] [t o t] ? a n s [t o t] = a [i] [j] (m o d 2)

$f[k][1]*ans[1]+f[k][2]*ans[2]+...f[k][tot]*ans[tot]=a[i][j] (mod 2)$

那末现在有 $tot$ 个方程， $tot$ 个未知数，高斯消元就能够直接出来了~

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#define N 5
#define M 6
using namespace std;
int a[35][35],ans[35],id[10][10],d[15][3];
void Gauss(int n,int m)
{
    for (int i=1;i<=n;i++)
    {
        int j;
        for (j=i;j<=n&&!a[i][j];j++);
        if (j==n+1) continue;
        if (i!=j)
            for (int k=i;k<=m;k++)
                swap(a[i][k],a[j][k]);
        for (int j=i+1;j<=n;j++)
            if (a[j][i])
                for (int k=i;k<=m;k++)
                    a[j][k]^=a[i][k];
    }
    for (int i=n;i;i--)
    {
        for (int j=n;j>i;j--)
            a[i][m]^=(a[i][j]*a[j][m]);
        ans[i]=a[i][m];
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    int cnt=0;
    for (int i=1;i<=N;i++)
        for (int j=1;j<=M;j++)
            id[i][j]=++cnt;
    d[0][1]=d[0][2]=0;
    d[1][1]=d[2][1]=0,d[1][2]=1,d[2][2]=-1;
    d[3][2]=d[4][2]=0,d[3][1]=1,d[4][1]=-1;
    for (int t=1;t<=T;t++)
    {
        memset(a,0,sizeof(a));
        printf("PUZZLE #%d
",t);
        for (int i=1;i<=N;i++)
            for (int j=1;j<=M;j++)
                for (int k=0;k<=4;k++)
                {
                    int p=id[i+d[k][1]][j+d[k][2]];
                    if (p) a[id[i][j]][p]=1;
                    else a[id[i][j]][p]=0;
                }
        for (int i=1;i<=N;i++)
            for (int j=1;j<=M;j++)
                scanf("%d",&a[id[i][j]][31]);
        Gauss(cnt,cnt+1);
        int now=0;
        for (int i=1;i<=N;i++)
        {
            printf("%d",ans[++now]);
            for (int j=2;j<=M;j++)
                printf(" %d",ans[++now]);
            printf("
");
        }
    }
    return 0;
}

这里写图片描述

感悟：

WA是由于没有清零a数组，致使1些本来是0的在以后变成1了。

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