2*1格子问题

废话少说，代码说话

#include <stdio.h>static double cnt[12][1<<11];static int trans[16384][2];int rows, cols, ntrans;/* there are ((sqrt(2)+1)^c - (sqrt(2)-1)^c) * (sqrt(2)+2) / 4 transitions * which is the solution to T_{c} = 2 * T_{c-1} + T_{c-2} */void backtrack (int n, int from, int to)// 这个函数产生row与row之间的递推数组{if (n > cols) return;if (n == cols){trans[ntrans][0] = from;trans[ntrans][1] = to;++ntrans;return;}backtrack (n+2, from<<2, to<<2);// from层跟to层分别向左移动两位，这样from层两个位置已经铺满，导致to层连着铺两块backtrack (n+1, from<<1, (to<<1)|1);// from层跟to层分别向左移动一位，这样from层一个位置已经铺满，导致to层不能铺backtrack (n+1, (from<<1)|1, to<<1);// from层跟to层分别向左移动一位，这样from层这个位置是空的，to层可以利用这个位置竖着铺}int main (){freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);int r,t;while (scanf("%d%d",&rows,&cols) && rows!=0 ){if (rows < cols) { t = rows; rows = cols ; cols = t; }/* calculate map of possible transitions by linear backtracking */ntrans = 0;backtrack (0, 0, 0);for (r=0 ; r<=rows ; r++)for (t=0 ; t<(1<<cols) ; t++)cnt[r][t] = 0;cnt[0][0] = 1;for (r=0 ; r<rows ; r++) /* the r topmost rows are already filled */ // 利用递推数组直接往下加for (t=0 ; t<ntrans ; t++) /* perform all transitions */cnt[r+1][trans[t][1]] += cnt[r][trans[t][0]];printf ("%.0f/n", cnt[rows][0]);}return 0;}