状压DP
Mondriaan's Dream
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Time Limit: 3000MS |
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Memory Limit: 65536K |
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Total Submissions: 9938 |
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Accepted: 5750 |
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
1
2
3
5
144
51205
Source
Ulm Local 2000
用0表示竖着放的木块的上半部分,单独的1表示竖着的木块的下半部分,两个连续的1表示横着放的木块.
最后一行铺满的状态就是全部为1
上下行的兼容性:
1:下面为0上面一定不能为0
2:下面为1.....如果上面为0,那么这是一个单独的1
如果上面为1,那么这应该是一个连续的1,检查下一位是否为1,及下一位的上面是否为1
#include <iostream> #include <cstdio> #include <cstring>
using namespace std;
long long int dp[14][1<<14];
int r,c;
bool isfirst(int x)
{
int ans=0;
for(int i=0;i<c;i++)
if((x>>i)&1) ans++;
if(ans%2==1) return false;
for(int i=0;i<c;i++)
{
if((x>>i)&1)
{
if((x>>(i+1))&1)
i++;
else
return false;
}
}
return true;
}
void getfirstline()
{
for(int i=0;i<(1<<c);i++)
{
if(isfirst(i))
{
dp[1]=1;
/*
for(int j=0;j<c;j++)
{
printf("%d ",(i>>j)&1);
}
putchar(10);
*/
}
}
}
bool isCool(int xia,int shang)
{
for(int i=0;i<c;i++)
{
if(((xia>>i)&1)==0)
{
if(((shang>>i)&1)==0) return false;
}
else
{
if(((shang>>i)&1)==0) continue;
else
{
if(i+1>=c) return false;
if(((xia>>(i+1))&1)==1)
{
if(((shang>>(i+1))&1)==1)
i++;
else
return false;
}
else
return false;
}
}
}
return true;
}
int main()
{
while(scanf("%d%d",&r,&c)!=EOF&&r&&c)
{
memset(dp,0,sizeof(dp));
if(r<c) swap(r,c);
getfirstline();
for(int i=2;i<=r;i++)
{
for(int j=0;j<(1<<c);j++)
{
for(int k=0;k<(1<<c);k++)
{
if(isCool(j,k))
{
dp[j]+=dp[i-1][k];
}
}
}
}
printf("%I64d\n",dp[(1<<c)-1]);
}
return 0;
}
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* This source code was highlighted by YcdoiT. ( style: Codeblocks )