状压dp poj2411 Mondriaan's Dream POJ - 2411

最新推荐文章于 2021-08-10 19:37:33 发布

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最新推荐文章于 2021-08-10 19:37:33 发布

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解决荷兰画家Piet Mondriaan的梦境问题，通过编程计算不同方式用2x1的小矩形填充任意大小的大矩形。采用深度优先搜索策略，递归地遍历所有可能的状态。

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Mondriaan's Dream

POJ - 2411

Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on, for all of his paper was filled with squares and rectangles), he dreamt of filling a large rectangle with small rectangles of width 2 and height 1 in varying ways.

Expert as he was in this material, he saw at a glance that he'll need a computer to calculate the number of ways to fill the large rectangle whose dimensions were integer values, as well. Help him, so that his dream won't turn into a nightmare! Input

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

Sample Input

Sample Output

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>

using namespace std;

long long   dp[2050][2050];//注意应该取值范围是2的幂，否则数组越界，一定要使用long long
 int m, n;
void dfs(int i, int j, int state, int ne)
{
    if(j == n)
    {
        dp[i+1][ne] += dp[i][state];
        return;
    }
    if(((1<<j)&state)> 0)
        dfs( i, j + 1, state, ne);
    if(((1 << j)&state) == 0)
        dfs( i, j + 1, state, ne|(1<<j));
    if(((1 << j)&state) == 0&&((1 << (j+1))&state) == 0&&j+ 1 < n)
        dfs( i, j + 2, state, ne);
        return;
}
int main()
{
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        if(!n&&!m)
            break;
        memset(dp,0,sizeof(dp));
        dp[1][0] = 1;
        for(int i = 1; i <= m; i ++){
        for(int j = 0; j < (1 << n); j ++){
                if(dp[i][j])
                {
                    dfs(i,0,j,0);
                }
        }
    }
    printf("%lld\n", dp[m+1][0]);
    }

    return 0;
}