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

Output

Sample Input

1 2
1 3
1 4
2 2
2 3
2 4
2 11
4 11
0 0

Sample Output

1
0
1
2
3
5
144
51205

Source

Solution:

比较恶心的状态压缩 DP。

竖放的骨牌两格均标为 1，横放的骨牌左格标为 0，右格标为 1。

将每列的状态压缩，按行转移即可。

能否转移的限制条件其实只有两个，但实现较为复杂:

1. 若前一列的某一行为 0，则这是一个横放的骨牌，本列的这一行必须为 1。
2. 若前一列的某一行为 1，且本列的这一行也为 1，则这是竖放的骨牌，前一列的下一行和本列的下一行必须均为 1。本行这一条件满足后，直接跳过下一行的检查。

由于限制条件在行数确定之后就已经确定，我们可以建立邻接表存储状态的前驱。

Code: O(2^2·min{w,h}(w+h)) [972K, 219MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;

int h, w, S;
long long dp[15][2100];

struct Arc{
	int np;
	Arc *nxt;
};

struct Prev_Recorder{
	Arc *V[2100], E[1000000];
	int tope;
	
	inline void clear() {tope = 0, memset(V, 0, sizeof(V));}
	
	inline void addedge(int u, int v) {E[++tope].np = v, E[tope].nxt = V[u], V[u] = &E[tope];}
	
	inline void build(const int &range){
		clear();
		int S = 1 << range;
		for(register int r = 0; r < S; r++)
			for(register int s = 0; s < S; s++){
				bool flag = 0;
				for(register int i = 0; i < range; i++)
					if((r & 1 << i) == 0){  // If i-th number of previous column is 0 (-)
						if((s & 1 << i) == 0){  // Then i-th number of current column must be 1
							flag = 1;
							break;
						}
					}
				for(register int i = 0; i < range; i++){
					if((r & 1 << i) && (s & 1 << i)){  // If i-th numbers of both previous column and current column are 1 (|)
						if(i == range - 1 || (r & 1 << i + 1) == 0 || (s & 1 << i + 1) == 0){  // Then (i+1)-th numbers of both must be 1
							flag = 1;
							break;
						}
						else i++;  // Jump over the next line
					}
				}
				if(!flag) addedge(s, r);
			}
	}
} prev;

int main(){
	while(scanf("%d%d", &h, &w) != EOF && h){
		if(h & 1 && w & 1){
			puts("0");
			continue;
		}
		if(h > w) swap(h, w);
		memset(dp, 0, sizeof(dp));
		prev.build(h), S = 1 << h;
		for(register int s = 0; s < S; s++){
			bool flag = 0;
			for(register int i = 0; i < h; i++)
				if((s & 1 << i)){
					if((s & 1 << i + 1) == 0){
						flag = 1;
						break;
					}
					else i++;
				}
			if(flag) dp[1][s] = 0;
			else dp[1][s] = 1;
		}
		for(register int i = 2; i <= w; i++)
			for(register int s = 0; s < S; s++)
				for(register Arc *ne = prev.V[s]; ne; ne = ne->nxt)
					dp[i][s] += dp[i - 1][ne->np];
		printf("%lld\n", dp[w][S - 1]);
	}
	return 0;
}