poj1222（高斯消元，开关问题）

地址：http://poj.org/problem?id=1222

EXTENDED LIGHTS OUT

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6305 Accepted: 4140

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

20 1 1 0 1 01 0 0 1 1 10 0 1 0 0 11 0 0 1 0 10 1 1 1 0 00 0 1 0 1 01 0 1 0 1 10 0 1 0 1 11 0 1 1 0 00 1 0 1 0 0

Sample Output

PUZZLE #11 0 1 0 0 11 1 0 1 0 10 0 1 0 1 11 0 0 1 0 00 1 0 0 0 0PUZZLE #21 0 0 1 1 11 1 0 0 0 00 0 0 1 0 01 1 0 1 0 11 0 1 1 0 1

题意：输入目标状态，问怎样操作才可以使全关的灯达到目标状态。

思路：开关问题的典型，直接高斯消元就可以，可以拿这题作为高斯消元的入门题。

代码：

#include<iostream>#include<queue>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;#define LL __int64#define col 30int num[40][40],ans[40];void getans(){    for(int i=0,m=0;i<col,m<col;m++){        int k=i;        for(int j=i+1;j<col;j++){  //将最大的前移            if(abs(num[j][m])>abs(num[k][m]))                k=j;        }        if(k!=i){            for(int j=m;j<=col;j++)                swap(num[i][j],num[k][j]);        }        if(!num[i][m]) continue;  //有可能存在中间一列全都为零        for(int j=i+1;j<col;j++){            if(!num[j][m]) continue;            for(k=m;k<=col;k++)                num[j][k]^=num[i][k];  //这里想明白了，因为我们操作恒置为1，所以操作状态为0的开关会使其变为1，反之操作状态为1的开关会使其变为0        }        i++;    }    for(int i=col-1;i>=0;i--){        ans[i]=num[i][col];        for(int j=i+1;j<col;j++)            ans[i]^=(num[i][j]&ans[j]);  //需要注意，这种求答案只适用于开关问题一类的题目    }}int main(){    int t,cas=1;    scanf("%d",&t);    while(t--){        memset(num,0,sizeof(num));        for(int i=0;i<5;i++)            for(int j=0;j<6;j++){                int s=i*6+j;                num[s][s]=1;                if(i>0) num[s-6][s]=1;                if(i<4) num[s+6][s]=1;                if(j>0) num[s-1][s]=1;                if(j<5) num[s+1][s]=1;            }        for(int i=0;i<col;i++)            scanf("%d",&num[i][col]);        getans();        printf("PUZZLE #%d",cas++);        for(int i=0;i<col;i++)            if(i%6==0)printf("\n%d",ans[i]);            else printf(" %d",ans[i]);        puts("");    }    return 0;}