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　

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

int OriginTile[16];
int Tiling[16];
int Res[16];

int GetBit(int & n, int i)
{
    return (n >> i) & 1;
}

void SetBit(int & n, int i, int d)
{
    if (d)
        n |= (1 << i);
    else
        n &= ~(1 << i);
}

void ReverseBit(int & n, int i)
{
    n ^= (1 << i);
}

int main()
{
    ios::sync_with_stdio(false);

    int M, N;
    cin >> M >> N;
    for (int i = 0; i < M; ++i)
    {
        for (int j = 0; j < N; ++j)
        {
            int t;
            cin >> t;
            SetBit(OriginTile[i], j, t);
        }
    }

    int times = 1 << N;
    for (int k = 0; k < times; ++k)
    {
        memcpy(Tiling, OriginTile, sizeof(OriginTile));
        Res[0] = k;

        for (int i = 0; i < M; ++i)
        {
            for (int j = 0; j < N; ++j)
            {
                if (GetBit(Res[i], j))
                {
                    if (j > 0)
                        ReverseBit(Tiling[i], j-1);
                    ReverseBit(Tiling[i], j);
                    if (j < N - 1)
                        ReverseBit(Tiling[i], j+1);
                }
            }
            if (i < M-1)
            {
                Tiling[i+1] ^= Res[i];
                Res[i+1] = Tiling[i];
            }
        }

        if (Tiling[M-1] == 0)
        {
            for (int i = 0; i < M; ++i)
            {
                for (int j = 0; j < N; ++j)
                {
                    cout << GetBit(Res[i], j);
                    if (j < N-1)
                        cout << ' ';
                }
                cout << endl;
            }

            return 0;
        }
    }

    cout << "IMPOSSIBLE" << endl;

    return 0;
}