POJ 2411 Mondriaan's Dream

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mob604756fb8908 2018-01-07 16:39:00

文章标签 插头dp POJ #include 状压dp i++ 文章分类 代码人生

Mondriaan's Dream

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 18359		Accepted: 10487

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.
POJ 2411 Mondriaan

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

插头dp模板题之：1*2的多米诺骨牌放满网格纸的方案数。
这个玩意有超级多变种，比如说：
1.不放满
2.不允许有贯穿网格的直线
3.矩阵版(行和列其中一个<=6另一个血大)
4.(骨牌形状的多样性)

当然这些本质都是一样的：插头dp。
这种题的共同特点就是不能像普通状压dp一样把整行整列当做状态的一部分，这样不如把
适合题目的轮廓线当成状态的一部分来的简单。

(留个坑，基于连通性的状压dp还完全不会(据说转移方案一般都是两位数写起来贼带劲hhhh))

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#define ll long long
using namespace std;
ll f[2][3000],n,m,ans;
int ci[30],now,nxt,to;

int main(){
    ci[0]=1;
    for(int i=1;i<=20;i++) ci[i]=ci[i-1]<<1;
    
    while(scanf("%lld%lld",&n,&m)==2&&n&&m){
        if(n>m) swap(n,m);
        now=ans=0,memset(f,0,sizeof(f));
        
        /*
        if(n==1){
            if(m&1) puts("0");
            else printf("%d\n",m>>1);
            continue;
        }
        */
        
        f[0][ci[n]-1]=1;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++){
                nxt=now^1;
                memset(f[nxt],0,sizeof(f[nxt]));
                
                for(int S=0;S<ci[n];S++){
                    if(!(S&ci[n-1])){
                        //竖着放 
                        to=(S<<1)|1;
                        f[nxt][to]+=f[now][S];
                    }else{
                        if(j&&!(S&1)){
                            //横着放 
                            to=((S<<1)|3)^ci[n];
                            f[nxt][to]+=f[now][S];
                        }
                        //不放 
                        f[nxt][(S<<1)^ci[n]]+=f[now][S];
                    }
                }
                
                now=nxt;
            }
        
        ans=f[now][ci[n]-1];
        printf("%lld\n",ans); 
    }
    
    return 0;
}