poj 2411 Mondriaan's Dream（状压DP）

最新推荐文章于 2020-08-29 19:48:49 发布

原创最新推荐文章于 2020-08-29 19:48:49 发布 · 526 阅读

1 ·

CC 4.0 BY-SA版权

状压DP 专栏收录该内容

4 篇文章

订阅专栏

探讨如何通过编程解决荷兰画家Piet Mondriaan的梦境问题：使用2×1的小矩形来填充各种尺寸的大矩形，并计算不同的填充方式数量。该问题采用动态规划的方法进行求解。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12232		Accepted: 7142

Description

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

Source

Ulm Local 2000

思路和（前一篇文章）SGU 131 一样，这种类型题目关键就是要学会如何去枚举状态，以及状态的转移。

#include <iostream>
#include <cstring>
#include <cstdio>
#define LL long long
using namespace std;
const int N=12;

LL dp[N][1<<N];
int n,m,len;

void dfs(int x,int y,int s1,int s2,int b1,int b2)
{
    if(y>m)
    {

        if(!b1 && !b2)  dp[x][s1]+=dp[x-1][s2];
        return ;
    }
    if(!b1 && !b2)  dfs(x,y+1,s1*2+1,s2*2,0,0);
    if(!b1)         dfs(x,y+1,s1*2+1,s2*2+1-b2,1,0);
    dfs(x,y+1,s1*2+b1,s2*2+1-b2,0,0);
}

void initial()
{
    len=1<<m;
    memset(dp,0,sizeof(dp));
    dp[0][len-1]=1;
}

void solve()
{
    for(int i=1;i<=n;i++)  dfs(i,1,0,0,0,0);
    printf("%lld\n",dp[n][len-1]);
}

int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        if(m==0 && n==0)  break;
        initial();
        solve();
    }
    return 0;
}