POJ 2411 Mondriaan's Dream 插头dp

题目链接：

http://poj.org/problem?id=2411

Mondriaan's Dream

Time Limit: 3000MS
Memory Limit: 65536K
#### 问题描述
> 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!
#### 输入
> 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 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.
####样例输入
> 1 2
> 1 3
> 1 4
> 2 2
> 2 3
> 2 4
> 2 11
> 4 11
> 0 0
>
####样例输出
> 1
> 0
> 1
> 2
> 3
> 5
> 144
> 51205

题意

用1*2和2*1的骨牌排满n*m的方格。

题解

插头dp入门题

参考：port

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-9;

const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=13;
LL dp[2][1<<maxn];
int n,m;

int main() {
    while(scf("%d%d",&n,&m)==2&&n){
        if(n<m) swap(n,m);

        int pre=0,cur=1;
        clr(dp[cur],0);
        dp[cur][0]=1;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                swap(pre,cur);
                clr(dp[cur],0);

                int p1=(1<<j),p2=(1<<(j+1));

                for(int k0=0;k0<(1<<(m+1));k0++){
                    int k=k0;
                    ///注意！换行的时候的处理
                    if(!j) k=k0<<1;

                    if((p1&k)&&!(p2&k)){
                        dp[cur][k^p1]+=dp[pre][k0];
                    }else if(!(p1&k)&&(p2&k)){
                        dp[cur][k^p2]+=dp[pre][k0];
                    }else if(!(p1&k)&&!(p2&k)){
                        if(j<m-1) dp[cur][k^p2]+=dp[pre][k0];
                        if(i<n-1) dp[cur][k^p1]+=dp[pre][k0];
                    }
                }
            }

        }

        prf("%lld\n",dp[cur][0]);

    }
    return 0;
}

//end-----------------------------------------------------------------------