POJ 2411 Mondriaan's Dream (状压+dfs)

梦中矩形填充

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解决荷兰画家皮特·蒙德里安梦境中的数学问题：如何用2x1的小矩形填充更大的矩形，并计算所有可能的填充方式。采用递归深度优先搜索策略，通过状态压缩实现高效计算。

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Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 12405		Accepted: 7239

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

思路来源； http://blog.youkuaiyun.com/bossup/article/details/9339039

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>


#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

typedef __int64 ll;

#define eps 1e-8

#define fre(i,a,b)  for(i = a; i <b; i++)
#define mem(t, v)   memset ((t) , v, sizeof(t))
#define sf(n)       scanf("%d", &n)
#define sff(a,b)    scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf          printf
#define bug         pf("Hi\n")

using namespace std;

#define INF 0x3f3f3f3f
#define N 1<<11

ll dp[N][N];

int n,m;

void dfs(int dep,int up,int down,int pos)
{
	if(pos>=m)
	{
	    dp[dep+1][down]+=dp[dep][up];
	    return ;
	}

	if(up&(1<<pos))
	{
		dfs(dep,up,down,pos+1);
		return ;
	}

	dfs(dep,up,down|(1<<pos),pos+1);
	if(pos<m-1&&!(up&(1<<(pos+1))))
		dfs(dep,up,down,pos+2);

}

int main()
{
	int i,j;
	while(sff(n,m),n+m)
	{
		if(n<m) swap(n,m);

		mem(dp,0);
		dp[0][0]=1;

		int len=1<<m;

		fre(i,0,n)
		 fre(j,0,len)
		  if(dp[i][j])
			dfs(i,j,0,0);

	pf("%I64d\n",dp[n][0]);

	}
	return 0;
}