poj1222 EXTENDED LIGHTS OUT 开灯问题

Problem I

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other)

Total Submission(s) : 3   Accepted Submission(s) : 3

Problem 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
《挑战程序设计竞赛》上的题目 就不翻译了
思路：首先，同一个格子反转两次就会恢复原状，所以多次反转是多余的。此外，反转的鸽子的集合相同的话，其次序是无关紧要的。
因为如果从上往下进行反转的话，第一行就有多种操作方案，不妨指定好最上面一行的反转方法。就可以从第二行进行朴素判断了。
第一行的反转方式就用类似 状态压缩的方式2^m种方法
#include <iostream>#include <algorithm>#include <string.h>#include <stdio.h>#include <cmath>#include <vector>#include <queue>#include <utility>using namespace std;int opt[5][6], title[5][6], flip[5][6];const int dir[5][2] = {0, 0, -1, 0, 0, 1, 1, 0, 0, -1};const int n = 5, m = 6;int get(int x, int y) {    int c = title[x][y];    for (int i = 0; i < 5; ++ i) {        int xx = x + dir[i][0];        int yy = y + dir[i][1];        if (xx >= 0 && xx < n && yy >= 0 && yy < m) {            c += flip[xx][yy];        }    }    return c % 2;}int calc() {    for (int i = 1; i < n; ++i) {        for (int j = 0; j < m; ++j) {            if (get(i - 1, j)) {                flip[i][j] = 1;            }        }    }        for (int i = 0; i < m; ++i) {        if (get(n - 1, i))            return -1;    }        int res = 0;        for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            res += flip[i][j];        }    }    return res;}void solve() {    int res = -1;        for (int i = 0; i < (1 << m); ++i) {        memset(flip, 0, sizeof(flip));                for (int j = 0; j < m; ++j) {            flip[0][m - j - 1] = (i >> j) & 1;        }                int num = calc();                if (num >= 0 && (res < 0 || res > num)) {            res = num;            memcpy(opt, flip, sizeof(flip));        }    }        for (int i = 0; i < n; ++i) {        for (int j = 0; j < m; ++j) {            cout << opt[i][j] << (i == m - 1 ? "" : " ");        }        cout << endl;    }}int main() {        //freopen("in.txt", "r", stdin);    int t;    cin >> t;    for (int i = 1; i <= t; ++i) {        for (int i = 0; i < n; ++i) {            for (int j = 0; j < m; ++j) {                cin >> title[i][j];            }        }                cout << "PUZZLE #" << i << endl;                solve();    }}