SGU 131 贴地砖类型 状态压缩DP

用dp[i][j]表示要贴第i层时，第i-1层的状态， 然后每一层dfs枚举一下情况即可
dfs(x, y, pre, now) x层数 y枚举的当前列 pre 前一层状态, now当前层状态


根据当前列y的这位上pre和now状态，确定要贴的图形， 注意贴完以后pre层必须满状态，在贴的过程中处理必须优先把pre层贴满。


状态转移如下图：.表示空，*表示满

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
#define LL long long
int n, m;
LL dp[11][1<<9];
int f(int a) {
	return 1<<a;
}
LL cnt;
int N;
void dfs(int x, int y, int pre, int now) {
	if(y >= m) {
		dp[x+1][now] += cnt;
		return;
	}
	if( pre&f(y) && now&f(y)) {
	    dfs(x, y+1, pre, now);
	    return;
	}
	if( !(pre&f(y)) && !(now&f(y)) ) {
	    dfs(x, y+1, pre|f(y), now|f(y)); //*
	    if(y+1 < m) {                    //*
            if( !(now&f(y+1)) )                         //*
                dfs(x, y+1, pre|f(y), now|f(y)|f(y+1)); //**

            if( !(pre&f(y+1)) )                         //**
                dfs(x, y+1, pre|f(y)|f(y+1), now|f(y)); //*

            if( !(pre&f(y+1)) && !(now&f(y+1)) )          //**
                dfs(x, y+1, pre|f(y)|f(y+1), now|f(y+1)); // *
	    }
	    return;
	}
	if( pre&f(y) && !(now&f(y)) ) {
	    if( y+1 < m && !(now&f(y+1)) )                 //*
            dfs(x, y+1, pre|f(y), now|f(y)|f(y+1));    //**

	    if( y+1 < m && !(pre&f(y+1)) && !(now&f(y+1)) )    // *
            dfs(x, y+1, pre|f(y+1), now|f(y)|f(y+1));      //**
	    dfs(x, y+1, pre, now); // 不放
	    return;
	}
	if( !(pre&f(y)) && now&f(y)) {
	    if(y+1 < m && !(pre&f(y+1) && !(now&f(y+1))) ) //**
            dfs(x, y+1, pre|f(y)|f(y+1), now|f(y+1));  // *
	}

}
int main() {
	int i, j;
	while(~scanf("%d%d", &n, &m)) {
		N = 1<<m;
		memset(dp, 0, sizeof(dp));
		dp[0][N-1] = 1;
		for(i = 0; i < n; i++)
			for(j = 0; j < N; j++) if(dp[i][j]) {
				cnt = dp[i][j];
				dfs(i, 0, j, 0);
			}
		cout << dp[n][N-1] << endl;
	}
}
/*
2 3
5

9 9
38896105985522272

6 6
5350806
*/