POJ 1222 && POJ 1830 开关问题 (高斯消元)

EXTENDED LIGHTS OUT

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7217		Accepted: 4724

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

 

最近想去学习下数学方面的知识,就先接触下高斯消元把...看bin神的blog一边刷题..

这里的关于高斯消元讲解满详细：http://www.cnblogs.com/iwtwiioi/p/4065711.html

 

题意:一个01矩阵,表示灯的亮灭状态,每次操作可以改变一个十字形状内的五个灯的状态。问能否将所有灯熄灭。

 

思路：高斯消元法。对于每个灯的两灭有影响的开关就是它附近十字形内的五个开关。所以对于每个灯可以列一个方程，即周围五个开关异或起来的结果应该可以使该灯熄灭。

就是利用线性代数知识，写出增广矩阵，化为阶梯形矩阵，有下到上依次解出各未知量。

 

代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;

//对2取模的01方程组
const int MAXN = 40;
//有equ个方程，var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元（多解枚举自由变元可以使用）
int free_num;//自由变元的个数

//返回值为-1表示无解，为0是唯一解，否则返回自由变元个数
int Gauss()
{
    int max_r,col,k;
    free_num = 0;
    for(k = 0, col = 0 ; k < equ && col < var ; k++, col++)
    {
        max_r = k;
        for(int i = k+1; i < equ; i++)
        {
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        }
        if(a[max_r][col] == 0)
        {
            k--;
            free_x[free_num++] = col;//这个是自由变元
            continue;
        }
        if(max_r != k)
        {
            for(int j = col; j < var+1; j++)
                swap(a[k][j],a[max_r][j]);
        }
        for(int i = k+1; i < equ; i++)
        {
            if(a[i][col] != 0)
            {
                for(int j = col; j < var+1; j++)
                    a[i][j] ^= a[k][j];
            }
        }
    }
    for(int i = k; i < equ; i++)
        if(a[i][col] != 0)
            return -1;//无解
    if(k < var) return var-k;//自由变元个数
    //唯一解，回代
    for(int i = var-1; i >= 0; i--)
    {
        x[i] = a[i][var];
        for(int j = i+1; j < var; j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}
void init()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
    equ = 30;
    var = 30;
    for(int i = 0; i < 5; i++)
        for(int j = 0; j < 6; j++)
        {
            int t = i*6+j;
            a[t][t] = 1;
            if(i > 0)a[(i-1)*6+j][t] = 1;
            if(i < 4)a[(i+1)*6+j][t] = 1;
            if(j > 0)a[i*6+j-1][t] = 1;
            if(j < 5)a[i*6+j+1][t] = 1;
        }
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T;
    int iCase = 0;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        init();
        for(int i = 0; i < 30; i++)
            scanf("%d",&a[i][30]);
        Gauss();
        printf("PUZZLE #%d\n",iCase);
        for(int i = 0; i < 5; i++)
        {
            for(int j = 0; j < 5; j++)
                printf("%d ",x[i*6+j]);
            printf("%d\n",x[i*6+5]);
        }
    }
    return 0;
}

 

附个POJ 1830代码

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
//对2取模的01方程组
const int MAXN = 40;
//有equ个方程，var个变元。增广矩阵行数为equ,列数为var+1,分别为0到var
int equ,var;
int a[MAXN][MAXN]; //增广矩阵
int x[MAXN]; //解集
int free_x[MAXN];//用来存储自由变元（多解枚举自由变元可以使用）
int free_num;//自由变元的个数

//返回值为-1表示无解，为0是唯一解，否则返回自由变元个数
int Gauss()
{
    int max_r,col,k;
    free_num = 0;
    for(k = 0, col = 0 ; k < equ && col < var ; k++, col++)
    {
        max_r = k;
        for(int i = k+1;i < equ;i++)
        {
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        }
        if(a[max_r][col] == 0)
        {
            k--;
            free_x[free_num++] = col;//这个是自由变元
            continue;
        }
        if(max_r != k)
        {
            for(int j = col; j < var+1; j++)
                swap(a[k][j],a[max_r][j]);
        }
        for(int i = k+1;i < equ;i++)
        {
            if(a[i][col] != 0)
            {
                for(int j = col;j < var+1;j++)
                    a[i][j] ^= a[k][j];
            }
        }
    }
    for(int i = k;i < equ;i++)
        if(a[i][col] != 0)
            return -1;//无解
    if(k < var) return var-k;//自由变元个数
    //唯一解，回代
    for(int i = var-1; i >= 0;i--)
    {
        x[i] = a[i][var];
        for(int j = i+1;j < var;j++)
            x[i] ^= (a[i][j] && x[j]);
    }
    return 0;
}
void init()
{
    memset(a,0,sizeof(a));
    memset(x,0,sizeof(x));
}
int start[MAXN],end[MAXN];

int main()
{
    //freopen("in.txt","r",stdin);
    int n,T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i = 0;i < n;i++)
            scanf("%d",&start[i]);
        for(int i = 0;i < n;i++)
            scanf("%d",&end[i]);
        init();
        equ = var = n;
        for(int i = 0;i < n;i++)
            a[i][i] = 1;
        int u,v;
        while(scanf("%d%d",&u,&v) == 2)
        {
            if(u == 0 && v == 0)break;
            a[v-1][u-1] = 1;
        }
        for(int i = 0;i < n;i++)
            a[i][n] = (start[i]^end[i]);
        int ans = Gauss();
        if(ans == -1)
            printf("Oh,it's impossible~!!\n");
        else printf("%d\n",(1<<ans));
    }
    return 0;
}