poj2411Mondriaan's Dream

Mondriaan's Dream

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 11810 Accepted: 6866

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. 

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 21 31 42 22 32 42 114 110 0

Sample Output

10123514451205

题意：给你n*m的地板，1*2的地板砖，问将地板铺满的方案数

解题思路：刚刚接触轮廓线dp，这题是比较经典的，1*2的地板砖可以横放，竖放，还可以不放，我们只需分三种情况讨论他的轮廓线状态，详细分析参见大白书。

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define LL long longLL dp[2][1<<11];int m;void update(int t,int a,int b){    if(b&(1<<m))    dp[t][b^(1<<m)]+=dp[1-t][a];}int main(){    int n;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0&&m==0)break;        int t=0;        memset(dp,0,sizeof(dp));        dp[0][(1<<m)-1]=1;        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                t=1-t;                memset(dp[t],0,sizeof(dp[t]));                for(int k=0;k<(1<<m);k++)                {                    if(!dp[1-t][k])continue;                    //bufang                    update(t,k,k<<1);                    //shufang                    if(i&&!((1<<m-1)&k))                    update(t,k,(k<<1)^1^(1<<m));                    //hengfang                    if(j&&!(k&1))                    update(t,k,(k<<1)^3);                }            }        }        printf("%lld\n",dp[t][(1<<m)-1]);    }    return 0;}