poj 2411 Mondriaan's Dream 状态压缩DP 打表

　　将题目转换题意, 有一个M列的墙, 我们需要 使用 1*2 的小矩形 砌成N层高, 有多少不同的方案数量.

　　因为只有1*2的砖头, 且只有竖立或者横着放.

　　那么我们规定 竖立放置的砖头属于 较高的一层, 且 当前点放置砖头则为1,否则为0

　　那么我们可以得出结论:

　　对于 I 层 状态 X , 若当前位置为0,则下层必定为1. 则意味着下层必定包含 t = (~X)& Mask 

　　对于 I+1 层 状态 Y, 除了包含 状态 t 之外, 剩下标记为1的 必须两两成对.才可符合状态.

　　

　　注意   t = (~X)& Mask    这里要记得还要与 总集合MASK相与. 把状态限定在MASK之内

解题代码:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef long long LL;
typedef unsigned int uint;
const int N = 1<<11;
/*
计算结果代码部分
LL ans[12][12];

LL dp[12][N+10];
int n, m;

bool check( uint x ){
    
    for(uint i = 0; i < m; i++)
    {
        if( x&(1<<i) )
        {
            if( i+1 >= m ) return false;
            if( ( x&(1<<(i+1) ) ) == 0 ) return false;
            i++;
        }
    }
    return true;
}

int main()
{
    freopen("6.out","w",stdout);

    memset( ans, 0, sizeof(ans) );
    for(n = 1; n <= 11; n++)
        for(m = 1; m <= 11; m++)
    {
        memset( dp, 0, sizeof(dp) );
    
        int Max = (1<<m)-1;
        for(int mask = 0; mask <= Max; mask++)
            if( check(mask) ) dp[0][mask] = 1;
    
    //    for(uint i = 0; i < Max; i++)
    //        if( dp[0][i] ) printf("%d ", i); puts("");
    
        for(int h = 0; h < n-1; h++)
        {
            for(uint mask = 0; mask <= Max; mask++)
            {
                if( dp[h][mask] ){
                    uint t = (~mask)&Max;
                    for(uint x = 0; x <= Max; x++)
                    {
                //        printf("mask = %u, t = %u, x = %u\n", mask, t, x );    
                        if( ((x&t)==t) && check( t^x ) ) 
                            dp[h+1][x] += dp[h][mask];    
                    
                    }    
                }    
            }
    //        cur = !cur;    
        }
    //    printf( (n==1&&m==1) ? "%lld" : ",%lld", dp[n-1][Max] );    
        ans[n][m] = dp[n-1][Max];
    }
    printf("{\n");
    for(int i = 0; i <= 11; i++)
    {
        printf("{");    
        for(int j = 0; j <= 11; j++)
        {
            printf( j == 0 ? "%lld" : ",%lld", ans[i][j] );    
        }
        printf( i == 11 ? "}\n" :  "},\n");        
    }
    printf("};\n");    
    return 0;
}
*/
LL ans[12][12] =
{
{0,0,0,0,0,0,0,0,0,0,0,0},
{0,0,1,0,1,0,1,0,1,0,1,0},
{0,1,2,3,5,8,13,21,34,55,89,144},
{0,0,3,0,11,0,41,0,153,0,571,0},
{0,1,5,11,36,95,281,781,2245,6336,18061,51205},
{0,0,8,0,95,0,1183,0,14824,0,185921,0},
{0,1,13,41,281,1183,6728,31529,167089,817991,4213133,21001799},
{0,0,21,0,781,0,31529,0,1292697,0,53175517,0},
{0,1,34,153,2245,14824,167089,1292697,12988816,108435745,1031151241,8940739824},
{0,0,55,0,6336,0,817991,0,108435745,0,14479521761,0},
{0,1,89,571,18061,185921,4213133,53175517,1031151241,14479521761,258584046368,3852472573499},
{0,0,144,0,51205,0,21001799,0,8940739824,0,3852472573499,0}
};

int main()
{
    int n, m;
    while( scanf("%d%d",&n,&m) != EOF)
    {
        if(n+m == 0 ) break;
        printf("%lld\n", ans[n][m]);
    }
    return 0;
}