poj2411 Mondriaan's Dream（轮廓线dp）

最新推荐文章于 2024-10-01 18:52:56 发布

zmy_wky

最新推荐文章于 2024-10-01 18:52:56 发布

阅读量220

点赞数

CC 4.0 BY-SA版权

分类专栏：动态规划文章标签：动态规划轮廓线dp

本文链接：https://blog.youkuaiyun.com/zmy_wky/article/details/79592246

动态规划专栏收录该内容

13 篇文章

订阅专栏

本文介绍使用轮廓线动态规划解决一道经典问题的方法，通过详细解析代码，展示如何高效地计算特定矩形区域中放置2×1尺寸矩形块的不同方式的数量。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

题目大意：给出一个 $n*m$ 的矩形，然后用 $1*2$ 大小的骨牌去填充 $n*m$ 的这个矩形，问有多少种填充方法。

今天做了一道轮廓线dp的题，发现轮廓线真的是妙啊，虽然代码短但是比其他dp题难理解，细节也很多。至于这道题的题解可以去看蓝书，然后这篇博客也写的不错：http://blog.youkuaiyun.com/u013480600/article/details/19499899

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

int n,m,cur;
long long dp[2][1<<13];

void update(int a,int b){
    if(b&(1<<m))dp[cur][b^(1<<m)]+=dp[cur^1][a];
}

int main()
{
    while(scanf("%d%d",&n,&m)){
        if(!n&&!m)break;
        memset(dp,0,sizeof(dp));
        cur=0;dp[0][(1<<m)-1]=1;
        for(int i=1;i<=n;i++)
          for(int j=1;j<=m;j++){
            cur^=1;
            memset(dp[cur],0,sizeof(dp[cur]));
            for(int k=0;k<(1<<m);k++){
                if(k&(1<<(m-1)))update(k,k<<1);
                if(i>1&&!(k&(1<<(m-1))))update(k,(k<<1)^1^(1<<m));
                if(j>1&&!(k&1)&&k&(1<<(m-1)))update(k,(k<<1)^3);
            }
          }
        printf("%lld\n",dp[cur][(1<<m)-1]);
    }
    return 0;
}