poj3279 Fliptile

思路：

枚举。

枚举了第一行的操作之后，下面每行的操作也随之确定了。因为在确定了第i行的操作之后，要想再改变a[i][j]的状态只能通过改变a[i + 1][j]来实现。另外，用到了集合的整数表示方法。

实现：

#include <iostream>
#include <cstring>
using namespace std;
const int INF = 0x3f3f3f3f;
const int dx[4] = {0, 1, 0, -1};
const int dy[4] = {1, 0, -1, 0};
int a[20][20], b[20][20], ans[20][20], buf[20][20], m, n;
void turn(int x, int y)
{
    b[x][y] = 1 - b[x][y];
    for (int i = 0; i < 4; i++)
    {
        int nx = x + dx[i], ny = y + dy[i];
        if (nx >= 0 && nx < m && ny >= 0 && ny < n)
        {
            b[nx][ny] = 1 - b[nx][ny];
        }
    }
}
int main()
{
    cin >> m >> n;
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
            cin >> a[i][j];
    }
    int minn = INF;
    for (int i = 0; i < (1 << n); i++)
    {
        int cnt = 0;
        memset(buf, 0, sizeof buf);
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                b[i][j] = a[i][j];
            }
        }
        for (int j = 0; j < n; j++)
        {
            int msk = 1 << j;
            if (i & msk)
            {
                turn(0, j); cnt++; buf[0][j] = 1;
            }
        }
        for (int j = 1; j < m; j++)
        {
            for (int k = 0; k < n; k++)
            {
                if (b[j - 1][k]) 
                {    
                    turn(j, k); cnt++; buf[j][k] = 1;
                }
            }
        }
        bool flg = true;
        for (int i = 0; i < n; i++)
        {
            if (b[m - 1][i]) { flg = false; break; }
        }
        if (flg && cnt < minn) 
        {
            minn = cnt;
            for (int i = 0; i < m; i++)
            {
                for (int j = 0; j < n; j++)
                {
                    ans[i][j] = buf[i][j];
                }
            }
        }
    }
    if (minn == INF) cout << "IMPOSSIBLE" << endl;
    else
    {
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
                cout << ans[i][j] << " ";
            cout << endl;
        }
    }
    return 0;
}