poj 2411 Mondriaan's Dream 狀態壓縮DP 打表

將題目轉換題意, 有一個M列的牆, 我們需要 使用 1*2 的小矩形 砌成N層高, 有多少不同的方案數量.

因為只有1*2的磚頭, 且只有豎立或者橫着放.

那么我們規定 豎立放置的磚頭屬於 較高的一層, 且 當前點放置磚頭則為1,否則為0

那么我們可以得出結論:

對於 I 層 狀態 X , 若當前位置為0,則下層必定為1. 則意味着下層必定包含 t = (~X)& Mask

對於 I+1 層 狀態 Y, 除了包含 狀態 t 之外, 剩下標記為1的 必須兩兩成對.才可符合狀態.

View Code
```#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef long long LL;
typedef unsigned int uint;
const int N = 1<<11;
/*

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(uint i = 0; i < Max; i++)
//        if( dp[0][i] ) printf("%d ", i); puts("");

for(int h = 0; h < n-1; h++)
{
{
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 ) )

}
}
}
//        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;
}```