zoj 1100

      這題是簡單的狀態壓縮，和黑書的那個bug公司類似，不過更簡單。代碼如下：（當然，可以打表）

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

#define HW	20
#define hpmax(a,b)	(a)>(b)?(a):(b)

long long	dp[2][2048];	//2^11
int factor[13] = { 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048 };
unsigned char state[13];
int h, w, p, q;

void parse( int s ) 
{
	int i;
	for ( i = 1; i <= w; ++ i ) {
		state[i] = s&1;
		s >>= 1;
	}
}

void dfs( int curr, long long cnt, int col )
{
	if ( col > w ) {
		if ( dp[q][curr] > 0 )
			dp[q][curr] += cnt;
		else
			dp[q][curr] = cnt;
		return;
	}

	/*vertical*/
	if ( 0 == state[col] ) {
		curr += factor[col-1];
		dfs( curr, cnt, col+1 );
	}

	/*horizon*/
	else {
		if ( 1 == state[col+1] && col < w ) {
			curr += factor[col-1] + factor[col];
			dfs( curr, cnt, col+2 );
			curr -= factor[col-1] + factor[col];
		}
		dfs( curr, cnt, col+1 );		
	}
}

int main()
{
	int i, j, curr;
	
	while ( scanf("%d%d", &h, &w ) != 0 && h && w ) {
		if ( w > h ) {
			curr = w; w = h; h = curr;
		}	
		memset( dp, -1, sizeof(dp) );
		dp[0][ factor[w]-1 ] = 1;
		p = 0;
		for ( i = 1; i <= h; ++ i ) {
			q = 1 - p;
			memset( dp[q], -1, sizeof(dp[q]) );
			for ( j = 0; j < factor[w]; ++ j ) {
				if ( -1 == dp[p][j] )
					continue;
				parse(j);
				curr = 0;	
				dfs( curr, dp[p][j], 1 );
			}
			p = q;
		}	
		if( dp[p][ factor[w]-1 ] < 0 )
			dp[p][ factor[w]-1 ] = 0; 
		printf("%lld\n", dp[p][factor[w]-1] );
	}
	return 0;
}