POJ2411 Mondriaan's Dream DP

Title:

A $n\ast m$ 's chessboard is filled with$1\ast 2$ small rectangles, how many kinds of schemes are there in total?

answer:

Getting started with DP. There are only 3 kinds of rectangles in each row:

1. Horizontal $1\ast 2$ rectangle

2. Vertical $1\ast 2$ rectangle, this row is above

3. Vertical $1\ast 2$ rectangle, this row is below

Simple analysis is available, use one $The\; m$ -bit binary number represents the placement of the rectangle in this row. Where thisbitis$0$ means nothing to do with the next line (1, 3),$1$ means related to the next line (2). $F\left[i\right]\left[j\right]$ means the$The$ placement of$i$ row isjjNumber of plans at$j$ .

对于$F[i-1]\left[k\right]$ transfer toF [i] [j] F[i][j]The conditions of$F$$[$$i$$]$$[$$j$$]$ are:

• $k\&j=0$ for$All$ below 1 in$k$ are$0$。
• $k|j$ each continuous0 0 in thebinary representation$0$ is an even number. $k|j$ is 1 representing a vertical rectangle, including both 2 and 3. The even number is guaranteed 1.

Which continuous 0 0 in the m-bit binary number can be preprocessed$0$ is an even number.

Error-prone:

Can't preprocess all at once, because continuous 0 0The number of$0$ is related to the number of digits! For example, when$m=7$ o'clock,$0$ is illegal; but when$m=At\; 8$ o'clock, 0 is legal. So readmmAccording tommafter$m$$m$ size preprocessing.

The preprocessing is also very clever. Use $odd$ indicates whether there is an odd number of consecutive$0$ ,$cnt$ means continuous$0$ is odd or even. If this bit is$1$，$odd|=cnt,cnt=0$ , there is no odd number of consecutive0 0 in theupdate memory$0$ , will$cnt$ initialization; if it is not 1, then$cn{t}^{\wedge }=1$ , which means the current segment of continuous0 0Whether$0$ is odd or even. After each bit of the binary is traversed, if the last segment is$0$ , thencnt cntat this time$cnt$ has not updated$odd$ , so$odd|cnt==When\; 1$ , it means there are an odd number of consecutive$0$ , otherwise it is legal.

Output $F\left[n\right]\left[0\right]$ , not$F[n][m]$。

AC code:

#include <cstdio>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define lep(i,a,b) for(int i=(a);i>=(b);i--) 
#define lepp(i,a,b) for(int i=(a);i>(b);i--)
#define pii pair<int,int>
#define pll pair<long long,long long>
#define mp make_pair
#define pb push_back 
#define fir first
#define sec second
#define All(x) x.begin(),x.end() 
#define ms(a,b) memset(a,b,sizeof(a)) 
#define INF 0x3f3f3f3f
#define INFF 0x3f3f3f3f3f3f3f3f
#define multi int T;scanf("%d",&T);while(T--) 
using namespace std;
typedef long long ll;
typedef double db;
const int N=(1<<11)+5;
const int mod=10007;
const db eps=1e-6;                                                                            
const db pi=acos(-1.0);
int n,m,zero2[N];
ll f[15][N];
int main(){
    #ifndef ONLINE_JUDGE
    freopen("D:\\work\\data.in","r",stdin);
    #endif
    // repp(i,0,1<<11){
    //     int cnt=0,odd=0;
    //     repp(j,0,11){
    //         if(i>>j&1) odd|=cnt,cnt=0;
    //         else cnt^=1;
    //     }
    //     zero2[i]=odd|cnt?0:1;
    // }
    f[0][0]=1;
    while(cin>>n>>m&&n){
        repp(i,0,1<<m){
            int cnt=0,odd=0;
            repp(j,0,m){
                if(i>>j&1) odd|=cnt,cnt=0;
                else cnt^=1;
            }
            zero2[i]=odd|cnt?0:1;
        }
        rep(i,1,n)
            repp(j,0,1<<m){
                f[i][j]=0;
                repp(k,0,1<<m){
                    if((j&k)==0&&zero2[j|k]) f[i][j]+=f[i-1][k];
                }
            }
        cout<<f[n][0]<<endl;
    }    
}