poj2411 Mondriaan's Dream

思路：

状压dp，只需记录m位状态即可。

实现：

#include <iostream>
#include <cstring>
using namespace std;
const int MAXN = 11, MAXM = 11;
typedef pair<int, int> pii;
typedef long long ll;
ll dp[MAXN + 1][MAXM + 1][1 << MAXM + 1];
int n, m;

pii next(int x, int y)
{
    pii res;
    res.first = y == m - 1 ? x + 1 : x;
    res.second = y == m - 1 ? 0 : y + 1;
    return res;
}

int main()
{
    while (cin >> n >> m, n || m)
    {
        memset(dp, 0, sizeof dp);
        dp[n][0][0] = 1;
        for (int i = n - 1; i >= 0; i--)
        {
            for (int j = m - 1; j >= 0; j--)
            {
                for (int k = 0; k < 1 << m; k++)
                {
                    pii tmp = next(i, j);
                    if (k >> j & 1)
                        dp[i][j][k] = dp[tmp.first][tmp.second][k & ~(1 << j)];
                    else
                    {
                        dp[i][j][k] = dp[tmp.first][tmp.second][k | 1 << j];
                        if (j < m - 1 && !(k >> j + 1 & 1))
                        {
                            tmp = next(i, j + 1);
                            dp[i][j][k] += dp[tmp.first][tmp.second][k];
                        }
                    }
                }
            }
        }
        cout << dp[0][0][0] << endl;
    }
    return 0;
}