POJ 2411 Mondriaan'sDream（状压DP）

G - Mondriaan's Dream

Time Limit:3000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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

【题目大意】一个矩阵，只能放1*2的木块，问将这个矩阵完全覆盖的不同放法有多少种。

【解析】如果是横着的就定义11，如果竖着的定义为竖着的01，这样按行dp只需要考虑两件事儿，当前行&上一行，是不是全为1，不是说明竖着有空（不可能出现竖着的00），另一个要检查当前行里有没有横放的，但为奇数的1。

原代码链接：http://blog.csdn.net/accry/article/details/6607703

首先我个人感觉，横着是11，竖着是01 这个方法很牛逼，然后就是先预处理ok数组，之后就要判断符合的情况，最后写dp方程，先写边界，再写之后。

之后枚举s，s1两种状态，判断s行，s1行 两行在一起是否可行，还有一个有意思的地方就是，中间的行数只要按位与==full-1就可以，但最后一行必须是全为full-1才行！！

#include <map>#include <set>#include <list>#include <cmath>#include <queue>#include <stack>#include <vector>#include <cstdio>#include <string>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int INF=0x3f3f3f3f;typedef long long ll;typedef unsigned long long ull;#define fi first#define se second#define prN printf("\n")#define SI(N) scanf("%d",&(N))#define SII(N,M) scanf("%d%d",&(N),&(M))#define SIII(N,M,K) scanf("%d%d%d",&(N),&(M),&(K))#define cle(a,val) memset(a,(val),sizeof(a))#define rep(i,b) for(int i=0;i<(b);i++)#define Rep(i,a,b) for(int i=(a);i<=(b);i++)int N,M;ll dp[1<<11][11];//注意这，状态数是第一维bool ok[1<<11];int full;//用来初始化ok数组，这个数组存着可行的状态（就是横着，并且有连着的两个1的时候，这就相当于放了一个横着的1*2木块，//剩下的0可以放竖着的，但x状态与y状态 按位“|”的时候，如果==full-1 那就可行 ）bool judge(int n){    int bit=0;    while(n)    {        if ((n&1))            bit++;        else        {            if ((bit&1))                return false;        }        n>>=1;    }    if ((bit&1))        return false;    return true;}bool judge2(int x,int y){    if ((x|y)!=full-1)//x与y  这两个状态必须要能完全覆盖两行才能继续进行        return false;    //还有可能是奇数的1的情况，所以要返回ok数组的值    return ok[(x&y)];}int main(){#ifndef ONLINE_JUDGE    freopen("C:\\Users\\Zmy\\Desktop\\in.txt","r",stdin);//    freopen("C:\\Users\\Zmy\\Desktop\\out.txt","w",stdout);#endif // ONLINE_JUDGE    full=1<<11;    rep(S,full)    if (judge(S))        ok[S]=1;    while(cin>>N>>M,N||M)    {        cle(dp,0);        full=1<<M;                //初始化dp边界        rep(S,full)        if(ok[S])            dp[S][0]=1;        //求dp 1到 n-1        Rep(i,1,N-1)        {            rep(s,full)            {                rep(s1,full)                {                    if (!judge2(s,s1)) continue;                    dp[s][i]+=dp[s1][i-1];//只有在都是1的情况下才做+=操作                }            }        }        //输出全是1的情况，并且在n-1行        int s=(1<<M)-1;        printf("%I64d\n",dp[s][N-1]);    }    return 0;}