A - Mondriaan's Dream （状态压缩DP）

Posted 2021-03-23 ylrwj

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

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

Sample Output

1
0
1
2
3
5
144
51205

题意：给你一个矩形的高和宽，让你用高为1宽为2的小矩形完全填充，不能重叠，不能超出，求出有多少种方案
如高为2，宽为4，有5种

 

这是状态压缩dp，https://blog.csdn.net/u014634338/article/details/50015825

将状态用二进制比特位表示，

 

横放为1，竖放对下一行产生影响，为01，如果最后一行全为1，则可以

#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cstdio>
using namespace std;
const int maxn=1e8;
typedef long long ll;
int n,m;
ll dp[15][1<<12];

bool init(int s)
{
    for(int j=0;j<m;)//前面j-1列符合要求，判断j列 
    {
        if(s&(1<<j))//第j列为1 
        {
            if(j==m-1)
              return false;
            if(s&(1<<(j+1)))//j和j+1列都为1，横放 
              j+=2;
            else//j为1，j+1为0不可行 
              return false;
        }
        else//j列为0，竖着放，可行 
          j++;
    }
    return true;
}
bool check(int w,int p)
{
    for(int j=0;j<m;)
    {
        if(w&(1<<j))//第i行j列为1 
        {
            if(p&(1<<j))//第i-1行j列也为1，第i行横放 
            {
                //所以第i行和i-1行的第j+1都为1，否则非法 
                if(j==m-1|| !(w&1<<(j+1))|| !(p&1<<(j+1)))
                   return false;
                else
                   j+=2;
            }
            else
              j++;
        }
        else
        {
            if(p&(1<<j))
              j++;
            else
              return false; 
        }
    }
    return true;
}
void solve()
{
    int t=(1<<m)-1;
    memset(dp,0,sizeof(dp));
    for(int i=0;i<=t;i++)
    {
        if(init(i))
           dp[1][i]=1;
    }
    for(int i=2;i<=n;i++)
    {
        for(int j=0;j<=t;j++)//第i行状态 
        {
            for(int k=0;k<=t;k++)//第i-1行的状态 
            {
                if(check(j,k))
                  dp[i][j]+=dp[i-1][k];
            }
        }
    }
    printf("%lld
",dp[n][t]); 
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        if(n==0||m==0)break;
        if(n&1 && m&1)//如果m，n都为奇数，面积为奇，不可能完全覆盖
        {
            printf("0
");
            continue;
        } 
        if(n<m)
          swap(n,m);
        solve();
    }
    return 0;
}