http://poj.org/problem?id=2411 import java.util.Scanner; import java.util.Arrays; public class Main{ /* f[i][j]表示第i行,方格排布为二进制数j(第k位上为1表示凸出一个格子,为0表示不凸出) 的方案数。用DFS进行状态转移。*/ static long f[][]=new long[12][2048]; static int n, m; static void dfs(int i, int j, int jj, int s)//j是初始状态,jj是目标状态.s表示列数 { if (s == m)//把i行m列放好 f[i + 1][jj] += f[i][j];//等于I+1行被占去的格子的2进制为JJ应该可以多放f[i][j]的方略 else if ((jj & (1 << s)) == 0)//表示第J列能放1/0 { dfs(i, j, jj | (1 << s), s + 1);//放1 if (s < m - 1 && (jj & (1 << (s + 1))) == 0) dfs(i, j, jj, s + 2);//放0(横占2格) } else//表示此处只能放0 dfs(i, j, jj & ~(1 << s), s + 1);//(jj & (1 << s)这个位置已经被占 } public static void main(String args[]){ Scanner sc=new Scanner(System.in); while (sc.hasNext()) { n=sc.nextInt(); m=sc.nextInt(); if(n+m==0) break; for(int i=0;i< f.length;i++) Arrays.fill(f[i],0); f[0][0] = 1; for (int i = 0; i < n; i++) for (int j = 0; j < (1 << m); j++) if (f[i][j]!=0) //剪枝(为0没有必要考虑) dfs(i, j, j, 0); System.out.printf("%dn", f[n][0]);//不占n行也即0~n-1放满的方法数 } } }
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