| Time Limit: 3000MS | Memory Limit: 65536K | |
| Total Submissions: 16061 | Accepted: 9297 |
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!
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
Source
动规 状压DP
这居然是状压DP……刚看题的时候满满都是插头DP的既视感 （后来知道插头DP也能做）
用01串代表一整行的状态，0表示格子暂时为空，1表示填满
转移时判断上一行状态和当前状态的每一位：
0 1
0 两格都为空，显然不行 0 可行，上一行填满，这一行留空等着放竖的
0 1
1 可行，上0下1表示一个竖块 1 如果下一列也是11，这代表两行各有一块横放。否则冲突，不可行
↑只有四个状态判断，多么简单（沙茶的我才用了足足一小时就想出来啦）
然后跑了900ms愉快垫底
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #define LL long long 7 using namespace std; 8 const int mxn=1<<13; 9 LL f[2][mxn]; 10 int now,pre; 11 int n,m; 12 bool check(int p,int r){ 13 // printf("p:%d r:%d ",p,r); 14 int tmp=1; 15 for(int i=0;i<m;i++){ 16 int x=p&tmp;int y=r&tmp; 17 if(!x && !y)return 0; 18 if((x && !y)||(!x && y)); 19 else if(x && y){ 20 tmp<<=1; 21 x=p&tmp;y=r&tmp; 22 if(x && y){i++;tmp<<=1;continue;} 23 else return 0; 24 } 25 tmp<<=1; 26 } 27 return 1; 28 } 29 int main(){ 30 int i,j; 31 while(scanf("%d%d",&n,&m) && n && m){ 32 if(n*m%2){ printf("%d ",0);continue;} 33 if(n<m)swap(n,m); 34 int ed=(1<<(m))-1; 35 now=0,pre=1; 36 memset(f,0,sizeof f); 37 for(i=0;i<=ed;i++){ if(check(ed,i))f[now][i]=1; } 38 for(i=2;i<=n;i++){ 39 swap(now,pre); 40 memset(f[now],0,sizeof f[now]); 41 for(j=0;j<=ed;j++){ 42 for(int k=0;k<=ed;k++){ 43 if(check(k,j))f[now][j]+=f[pre][k]; 44 } 45 } 46 } 47 printf("%lld ",f[now][ed]); 48 } 49 return 0; 50 }
每次状压写出来都跑的特别慢，我的常数可能是假的
之后看了别人的题解，发现可以根据上述转移推出更多性质而快速判定
比如说这样 （45、46行） 100ms
再深一步，用DFS预处理状态，可以优化到0ms，不过懒得写了233
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 #define LL long long 7 using namespace std; 8 const int mxn=1<<11; 9 LL f[2][mxn]; 10 bool ok[mxn]; 11 int now,pre; 12 int n,m; 13 bool check(int p,int r){ 14 // printf("p:%d r:%d ",p,r); 15 int tmp=1; 16 for(int i=0;i<m;i++){ 17 int x=p&tmp;int y=r&tmp; 18 if(!x && !y)return 0; 19 if((x && !y)||(!x && y)); 20 else if(x && y){ 21 tmp<<=1; 22 x=p&tmp;y=r&tmp; 23 if(x && y){i++;tmp<<=1;continue;} 24 else return 0; 25 } 26 tmp<<=1; 27 } 28 return 1; 29 } 30 int main(){ 31 int i,j; 32 while(scanf("%d%d",&n,&m) && n && m){ 33 if(n*m%2){ printf("%d ",0);continue;} 34 if(n<m)swap(n,m); 35 int ed=(1<<(m))-1; 36 now=0,pre=1; 37 memset(ok,0,sizeof ok); 38 memset(f,0,sizeof f); 39 for(i=0;i<=ed;i++){ if(check(ed,i))f[now][i]=1,ok[i]=1; } 40 for(i=2;i<=n;i++){ 41 swap(now,pre); 42 memset(f[now],0,sizeof f[now]); 43 for(j=0;j<=ed;j++){ 44 for(int k=0;k<=ed;k++){ 45 if((k|j)!=(1<<m)-1)continue; 46 if(!ok[k&j])continue; 47 if(check(k,j))f[now][j]+=f[pre][k]; 48 } 49 } 50 } 51 printf("%lld ",f[now][ed]); 52 } 53 return 0; 54 }