poj2411 Mondriaan's Dream 状态压缩dp

Mondriaan's Dream

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 11782 Accepted: 6852

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

Source

Ulm Local 2000

题意：用1*2的格子填满n*m的网格，求方法数。记得当初这题卡了我好久，看了各路神牛的题解也无从下手，后来终于懂了，还挺有成就感的。

思路：自己总结了一下，思路都是用0和1表示 我是用11表示横放 上一行0 下一行1表示竖放，因为这一行的状态只和上一行有关。这样表示最后一行一定都是1，结果就是dp[n][(1<<m)-1]，用dfs搜一遍所有情况，s1是上一行 s2表示下一行 用vector存一下，特判一下第一行，因为是第一行所以可以有任意个0，但相邻1的个数一定是偶数，然后dp搞一遍 做法不是最好的 但还算简洁明了吧。

注：这题也可以用插头dp的思路去做，写在我的另一篇博客http://blog.csdn.net/smz436487/article/details/38116053

这里就不另作说明了。

#include <iostream>#include <vector>#include <cstring>#include <cstdio>using namespace std;const int N=1<<11;long long int dp[2][N];vector<int> vec[N];void dfs(int cnt, int s1,int s2,int n){    if(n==cnt){         vec[s2].push_back(s1);         return ;     }     dfs(cnt+1,s1<<1,s2<<1|1,n);     dfs(cnt+1,s1<<1|1,s2<<1,n);//除了第三种情况 第一行是0的话，第二行一定是1，第一行是1，那第二行就是0，不明白自己画画推一推。     if(cnt<n-1)     dfs(cnt+2,s1<<2|3,s2<<2|3,n);//两行都是横放的情况}int pdf(int a)//特判第一行的情况 如果有奇数个相邻的1，就返回0，反之返回1{    while(a>0){        if((a&3)==3)a>>=2;        else if(a%2==0)a>>=1;        else return 0;    }    return 1;}int main(){    int n,m;    while(scanf("%d%d",&n,&m),n){        memset(dp[0],0,sizeof(dp[0]));        memset(vec,0,sizeof(vec));        dfs(0,0,0,m);        for(int i=0;i<1<<m;i++)            if(pdf(i))dp[0][i]=1;        for(int i=1;i<n;i++){            memset(dp[i&1],0,sizeof(dp[i&1]));            for(int j=0;j<1<<m;j++)                for(int k=0;k<vec[j].size();k++)                    dp[i&1][j]+=dp[i&1^1][vec[j][k]];        }        printf("%I64d\n",dp[n&1^1][(1<<m)-1]);    }    return 0;}