POJ 3279 Fliptile

题目大意:（熄灯问题），就是把1全变为0，变的时候附近的都要变

解题思路：我用的枚举，建议用高斯消元，这道题Kuangbin划分在简单搜索里的，搜索应该也能过

注：有IMPOSSIBLE的情况，开始交的时候忘考虑了　

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <algorithm>
 5 #include <cmath>
 6 using namespace std;
 7 
 8 int OriginTile[16];
 9 int Tiling[16];
10 int Res[16];
11 
12 int GetBit(int & n, int i)
13 {
14     return (n >> i) & 1;
15 }
16 
17 void SetBit(int & n, int i, int d)
18 {
19     if (d)
20         n |= (1 << i);
21     else
22         n &= ~(1 << i);
23 }
24 
25 void ReverseBit(int & n, int i)
26 {
27     n ^= (1 << i);
28 }
29 
30 int main()
31 {
32     ios::sync_with_stdio(false);
33 
34     int M, N;
35     cin >> M >> N;
36     for (int i = 0; i < M; ++i)
37     {
38         for (int j = 0; j < N; ++j)
39         {
40             int t;
41             cin >> t;
42             SetBit(OriginTile[i], j, t);
43         }
44     }
45 
46     int times = 1 << N;
47     for (int k = 0; k < times; ++k)
48     {
49         memcpy(Tiling, OriginTile, sizeof(OriginTile));
50         Res[0] = k;
51 
52         for (int i = 0; i < M; ++i)
53         {
54             for (int j = 0; j < N; ++j)
55             {
56                 if (GetBit(Res[i], j))
57                 {
58                     if (j > 0)
59                         ReverseBit(Tiling[i], j-1);
60                     ReverseBit(Tiling[i], j);
61                     if (j < N - 1)
62                         ReverseBit(Tiling[i], j+1);
63                 }
64             }
65             if (i < M-1)
66             {
67                 Tiling[i+1] ^= Res[i];
68                 Res[i+1] = Tiling[i];
69             }
70         }
71 
72         if (Tiling[M-1] == 0)
73         {
74             for (int i = 0; i < M; ++i)
75             {
76                 for (int j = 0; j < N; ++j)
77                 {
78                     cout << GetBit(Res[i], j);
79                     if (j < N-1)
80                         cout << ' ';
81                 }
82                 cout << endl;
83             }
84 
85             return 0;
86         }
87     }
88 
89     cout << "IMPOSSIBLE" << endl;
90 
91     return 0;
92 }