POJ - 1222 EXTENDED LIGHTS OUT(反转问题)

EXTENDED LIGHTS OUT

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9081 Accepted: 5904

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:有一个5*6的矩阵，每个位置都表示按钮和灯，1表示亮，0表示灭。每当按下一个位置的按钮，它和它周围灯的状态全部翻转，问在这样的一个方阵中按下哪些按钮可以把整个方阵都变成灭的，这时1表示按了，0表示没按。

按照题目的意思，是可能有多个解的，但是我们只需要输出其中一个可行解就可以了，而且提供的数据最少会存在一个可行解

思路:直接用区间反转的方法进行反转记录,针对最上方的点构造图形，原因:因为它只会被处理一遍，具体看代码(此题还可以用高斯消元法做)

/*头文件模板*/#include <map>#include <set>#include <cmath>#include <ctime>#include <queue>#include <vector>#include <cctype>#include <cstdio>#include <string>#include <cstring>#include <sstream>#include <cstdlib>#include <typeinfo>#include <iostream>#include <algorithm>#include <functional>using namespace std;#define pb push_back#define mp make_pair#define mem(a, x) memset(a, x, sizeof(a))#define copy(a, b) memcpy(a, b, sizeof(a))#define lson rt << 1, l, mid#define rson rt << 1|1, mid + 1, r#define FIN freopen("input.txt", "r", stdin)#define FOUT freopen("output.txt", "w", stdout)typedef long long LL;typedef pair<int, int > PII;typedef pair<int,string> PIS;typedef unsigned long long uLL;template<typename T>void print(T* p, T* q, string Gap = " ", bool flag = false) {int d = p < q ? 1 : -1;while(p != q) {if(flag) cout << Gap[0] << *p << Gap[1];else cout << *p;p += d;if(p != q && !flag) cout << Gap;}cout << endl;}template<typename T>void print(const T &a, string bes = "") {int len = bes.length();if(len >= 2)cout << bes[0] << a << bes[1] << endl;else cout << a << endl;}void IO_Init() {ios::sync_with_stdio(false);}LL LLabs(LL a) {return a > 0 ? a : -a;}const double PI = 3.1415926535898;const double eps = 1e-10;const int MAXM = 1e4 + 5;const int MAXN = 10 + 5;const LL INF = 0x3f3f3f3f;/*头文件模板*/int A[MAXN][MAXN];int F[MAXN][MAXN];//记录某个点在此之前被翻了几次int J[MAXN][MAXN];//记录哪些会被翻int dx[4] = {1,2,1,1};int dy[4] = {0,0,1,-1};/*针对最上方的点构造图形，原因:因为它只会被处理一遍 **** **/int T, res;void cal(int x, int y) {for(int i = 0; i < 4; i ++) {int nx = x + dx[i];int ny = y + dy[i];if(nx <= 0 || ny <= 0 || nx > 5 || ny > 6) continue;F[nx][ny] ^= 1;}}bool calc(int x) {memset(J, 0, sizeof(J));memset(F, 0, sizeof(F));for(int i = 0; i < 6; i ++) {A[0][i + 1] = (x & (1 << i)) > 0;}for(int i = 0; i < 5; i ++) {for(int j = 1; j <= 6; j ++) {if((F[i][j] + A[i][j]) & 1) {                F[i][j] = 1;//此处可有可无cal(i, j);J[i + 1][j] = 1;}}}for(int j = 1; j <= 6; j ++) {if((F[5][j] + A[5][j]) & 1) {return false;}}return true;}int main() {//FIN;int cas = 1;scanf("%d", &T);while(T --) {res = 0;memset(A, 0, sizeof(A));for(int i = 1; i <= 5; i ++) {for(int j = 1; j <= 6; j ++) {scanf("%d", &A[i][j]);}}for(int i = 0; i < (1 << 6); i ++) {if(calc(i)) {break;}}printf("PUZZLE #%d\n", cas ++);for(int i = 1; i <= 5; i ++) {for(int j = 1; j <= 6; j ++) {printf("%d%c", J[i][j], j == 6 ? '\n' : ' ');}}}return 0;}