POJ1222(SummerTrainingDay01-E)

Posted Penn000

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ1222(SummerTrainingDay01-E)相关的知识,希望对你有一定的参考价值。

EXTENDED LIGHTS OUT

Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 11078   Accepted: 7074

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. 
技术分享

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. 
技术分享

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

 
 1 //2017-08-01
 2 #include <cstdio>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 
10 bool grid[20][20], tmp[20][20];
11 int n, m, ans[20], res[20];
12 
13 void flip(int x, int y)
14 {
15     tmp[x][y] = !tmp[x][y];
16     if(x-1>=0)tmp[x-1][y] = !tmp[x-1][y];
17     if(y-1>=0)tmp[x][y-1] = !tmp[x][y-1];
18     tmp[x+1][y] = !tmp[x+1][y];
19     tmp[x][y+1] = !tmp[x][y+1];
20 }
21 
22 void solve()
23 {
24     int penn, minflip = 0x3f3f3f3f;
25     bool fg = false;
26     for(int i = 0; i < (1<<m); i++)
27     {
28         for(int x = 0; x < n; x++)
29               for(int y = 0; y < m; y++)
30                   tmp[x][y] = grid[x][y];
31         for(int y = 0; y < m; y++)
32             if(i&(1<<y))
33                   flip(0, m-1-y);
34         ans[0] = i;
35         for(int x = 1; x < n; x++){
36             penn = 0;
37             for(int y = 0; y < m; y++){
38                 if(tmp[x-1][y]){
39                     flip(x, y);
40                     penn += (1<<(m-1-y));
41                 }
42             }
43             ans[x] = penn;
44         }
45         bool ok = true;
46         for(int j = 0; j < m; j++)
47               if(tmp[n-1][j])
48                   ok = false;
49         if(ok){
50             fg = true;
51             int cnt = 0;
52             for(int j = 0; j < n; j++){
53                 for(int pos = 0; pos < m; pos++)
54                       if(ans[j]&(1<<(m-1-pos)))cnt++;
55             }
56             if(cnt < minflip){
57                 minflip = cnt;
58                 for(int k = 0; k < n; k++)
59                       res[k] = ans[k];
60             }
61         }
62     }
63     if(!fg)cout<<"IMPOSSIBLE"<<endl;
64     else{
65         for(int j = 0; j < n; j++){
66             for(int pos = 0; pos < m; pos++)
67                   if(pos == m-1)cout<<(res[j]&(1<<(m-1-pos))?1:0)<<endl;
68                 else cout<<(res[j]&(1<<(m-1-pos))?1:0)<<" ";
69         }
70     }
71 }
72 
73 int main(){
74     int T;
75     cin>>T;
76     for(int kase = 1; kase <= T; kase++){
77         n = 5;
78         m = 6;
79         for(int i = 0; i < n; i++)
80               for(int j = 0; j < m; j++)
81                   cin>>grid[i][j];
82         cout<<"PUZZLE #"<<kase<<endl;;
83         solve();
84     }
85 
86     return 0;
87 }

 










以上是关于POJ1222(SummerTrainingDay01-E)的主要内容,如果未能解决你的问题,请参考以下文章

POJ2478(SummerTrainingDay04-E 欧拉函数)

POJ3090(SummerTrainingDay04-M 欧拉函数)

POJ2411(SummerTrainingDay02-I 状态压缩dp)

poj1222 开关问题

POJ 1222 反转

POJ 1222 EXTENDED LIGHTS OUT