poj2411

http://www.elijahqi.win/2017/07/07/poj2411/

题目大意：

给出宽为2 高为1的长方形

接下来n行输入数据读入 h w 直到 0 0结束

求出一共有多少种方法可以使单位原始长方形填充满h w的矩形

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

The input contains several test cases. Each test case is made up of two integer numbers: the height h and the width w of the large rectangle. Input is terminated by h=w=0. Otherwise, 1<=h,w<=11.

Output

For each test case, output the number of different ways the given rectangle can be filled with small rectangles of size 2 times 1. Assume the given large rectangle is oriented, i.e. count symmetrical tilings multiple times.

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

#include<cstdio>
int state,h,w;
long long f[20][3000];
void dfs1(int i,int p1,int p2,int x){//计算第i行p2状态准备放置x位置 前一行为p1的方案数
/*
if x>w 表示所有列都放完  第i行得到新的状态p2 f[i][p2]+=f[i-1][p1] 第二件事 返回
if  p1的x位置==0；必须竖放  dfs1（i,p1,p2|1<<x-1,x+1);否则x位置==1{
x位置可以不放或者横放
不放：dfs1(i,p1,p2,x+1);
准备横放：if (x<w)不出界并且p1的x+1位置不为0 才可以横放
dfs(i,p1,p2+(修改x，x+1两个位置)，x+2）；
}
*/
if (x>w){
f[i][p2]+=f[i-1][p1];return ;
}
if ((p1&(1<<x-1))==0) dfs1(i,p1,p2+(1<<x-1),x+1);else{
dfs1(i,p1,p2,x+1);
if (x<w&&((p1&(1<<x))!=0)){
dfs1(i,p1,p2+(1<<x-1)+(1<<x),x+2);
}
}
}
int main(){
freopen("g1413.in","r",stdin);
freopen("g1413.out","w",stdout);
while (1){
scanf("%d%d",&h,&w);
if (h==0&&w==0) break;
state=(1<<w)-1;
for (int j=0;j<=state;++j) f[0][j]=0;
f[0][state]=1;

for (int i=1;i<=h;++i){
for (int j=0;j<=state;++j) f[i][j]=0;
for (int j=0;j<=state;++j){
if (f[i-1][j]!=0){
dfs1(i,j,0,1);//四个参数i表示准备放置第i行 i-1行状态是j i行状态从0开始且从第一个位置开始
}else continue;
}

}
printf("%I64d\n",f[h][state]);
}

return 0;
}