sgu131 Hardwood floor

最新推荐文章于 2019-09-04 18:01:00 发布

smilewsw

最新推荐文章于 2019-09-04 18:01:00 发布

阅读量695

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM题解文章标签： dp

本文链接：https://blog.youkuaiyun.com/smilewsw/article/details/40119705

ACM题解专栏收录该内容

35 篇文章

订阅专栏

131. Hardwood floor

time limit per test: 0.25 sec.
memory limit per test: 4096 KB

The banquet hall of Computer Scientists' Palace has a rectangular form of the size M x N (1<=M<=9, 1<=N<=9). It is necessary to lay hardwood floors in the hall. There are wood pieces of two forms:
1) rectangles (2x1)
2) corners (squares 2x2 without one 1x1 square)
You have to determine X - the number of ways to cover the banquet hall.
Remarks. The number of pieces is large enough. It is not allowed to leave empty places, or to cover any part of a surface twice, or to saw pieces.

Input

The first line contains natural number M. The second line contains a natural number N.

Output

First line should contain the number X, or 0 if there are no solutions.

Sample Input

2 3

Sample Output

轮廓线DP。由于这两种放法会占用下一填充块的下方位置。所以边界需将当前填充块的下方考虑进去。压缩状态时加一位，表示该位置。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
long long  dp[2][1500];
int main()
{
    int m,n,cnt,i,j,k;
    scanf("%d%d",&m,&n);
    memset(dp,0,sizeof(dp));
    cnt=0;
    dp[0][0]=1;
    for(i=0;i<m;i++)
    {
        for(j=0;j<n;j++)
        {
            memset(dp[cnt^1],0,sizeof(dp[cnt^1]));
            for(k=0;k<1<<(n+1);k++)
            {
                if(dp[cnt][k]==0)   continue;
                if(k>>j&1)
                {
                    if(k>>n&1)
                        dp[cnt^1][k&~(1<<n)]+=dp[cnt][k];
                    else
                        dp[cnt^1][k&~(1<<j)]+=dp[cnt][k];
                }
                else
                {
                    //1*2砖块
                    if(j<n-1&&!(k>>(j+1)&1))
                    {
                        if(k>>n&1)
                           dp[cnt^1][k&(~(1<<n))|(1<<(j+1))|1<<j]+=dp[cnt][k];

                        else
                            dp[cnt^1][k|(1<<(j+1))]+=dp[cnt][k];
                    }
                    if(i<m-1&&!(k>>n&1))
                        dp[cnt^1][k|(1<<j)]+=dp[cnt][k];

                    //3砖块
                    if(j<n-1&&i<m-1&&!(k>>(j+1)&1)&&!(k>>n&1))
                        dp[cnt^1][k|(1<<j)|(1<<(j+1))]+=dp[cnt][k];

                    if(j<n-1&&i<m-1&&!(k>>n&1))
                        dp[cnt^1][k|(1<<j)|(1<<n)]+=dp[cnt][k];

                    if(j>0&&i<m-1&&!(k>>(j-1)&1)&&!(k>>n&1))
                        dp[cnt^1][k|(1<<(j-1))|(1<<j)]+=dp[cnt][k];

                    if(j<n-1&&i<m-1&&!(k>>(j+1)&1))
                    {
                        if(k>>n&1)
                            dp[cnt^1][k|(1<<j)|1<<(j+1)]+=dp[cnt][k];
                        else
                            dp[cnt^1][k|(1<<(j+1)|(1<<n))]+=dp[cnt][k];
                    }
                }
            }
            cnt^=1;
        }
    }
    printf("%lld\n",dp[cnt][0]);
    return 0;
}