[状压dp] hdu 4064 Carcassonne

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=4064

Carcassonne

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 857    Accepted Submission(s): 326

Problem Description

Carcassonne is a tile-based board game for two to five players.

Square tiles are printed by city segments,road segments and field segments. 




The rule of the game is to put the tiles alternately. Two tiles share one edge should exactly connect to each other, that is, city segments should be linked to city segments, road to road, and field to field. 




To simplify the problem, we only consider putting tiles:

Given n*m tiles. You can rotate each tile, but not flip top to bottom, and not change their order. 

How many ways could you rotate them to make them follow the rules mentioned above?

 

Input

The first line is a number T(1<=T<=50), represents the number of case. The next T blocks follow each indicates a case.

Each case starts with two number N,M(0<N,M<=12)

Then N*M lines follow,each line contains M four-character clockwise.

'C' indicate City.

'R' indicate Road.

'F' indicate Field.

 

Output

For each case, output the number of ways mod 1,000,000,007.(as shown in the sample output)

 

Sample Input

3
1 1
RRRR
1 2
RRRF FCCC
8 8
FCFF RRFC FRCR FRFR RCCR FFCC RRFF CRFR
FRRC FRFR CCCR FCFC CRRC CRRR FRCR FRFR
RRCR FRRR CCCR FFFC RRFF RFCR CCFF FCCC
CFCF RRFF CRFR FFRR FRRF CCRR FFFC CRRF
CFRR FFFF FFFF RRFF RRRR RCRR FFCC RFRF
RRCF FRFR FRRR FRFR RCCR RCCC CFFC RFRF
CFCF FRFF RRFF FFFF CFFF CFFF FRFF RFRR
CCRR FCFC FCCC FCCC FFCC FCCF FFCC RFRF

 

Sample Output

Case 1: 4
Case 2: 1
Case 3: 1048576

 

Source

The 36th ACM/ICPC Asia Regional
Fuzhou Site —— Online Contest

 

Recommend

lcy

 

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题目意思：

每一个格子有四条边，每条边有一种颜色，求通过旋转格子，使相邻格子公共边颜色同样，总的种数。

解题思路：

状态压缩dp

格子不多仅仅有12个,对于每一行维护两种状态，上边的颜色状态up和下边的颜色状态dw，一行搜完后，用上一行的dw状态（也就是当前行的up状态）更新当前行的dw状态。

滚动数组处理。

剪枝：当一个格子四边都是一样颜色的话直接*4，不用再枚举那一条边在上面

代码：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 15
char sa[Maxn][Maxn][5];
ll dp[2][1100000];
int ba[Maxn],n,m,cur,mul;
map<char,int>myp;

void dfs(int r,int nu,int up,int dw,int ri)
{
    if(nu>m)
    {
        if(r==1)
        {
            dp[cur][dw]+=mul;
            dp[cur][dw]%=M;
        }
        else
        {
            dp[cur][dw]+=(dp[cur^1][up]*mul)%M;
            dp[cur][dw]%=M;
        }
        return ;
    }
    int i;
    for(i=1;i<4;i++)
        if(sa[r][nu][i]!=sa[r][nu][0])
            break;
   if(i==4) //四个面都同样
   {
       mul*=4;
       int a=myp[sa[r][nu][0]];
       if(ri==-1||a==ri)
            dfs(r,nu+1,up*3+a,dw*3+a,a);
       mul/=4;
       return ;
   }
   for(int i=0;i<4;i++)
   {
       int uu=myp[sa[r][nu][i]];
       int rr=myp[sa[r][nu][(i+1)%4]];
       int dd=myp[sa[r][nu][(i+2)%4]];
       int L=myp[sa[r][nu][(i+3)%4]];

       if(ri==-1||L==ri)
            dfs(r,nu+1,up*3+uu,dw*3+dd,rr);
   }
}
int main()
{
   //cout<<pow(3.0,12.0);

    //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   int t,cas=0;

   myp['F']=0;
   myp['R']=1;
   myp['C']=2;

   ba[0]=1;
   for(int i=1;i<=12;i++)
        ba[i]=ba[i-1]*3;

   scanf("%d",&t);
   while(t--)
   {
       scanf("%d%d",&n,&m);
       for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%s",sa[i][j]);
       memset(dp,0,sizeof(dp));
       cur=0;
       for(int i=1;i<=n;i++)
       {
           cur=cur^1;
           mul=1;
           dfs(i,1,0,0,-1);
           for(int j=0;j<ba[m];j++)
                dp[cur^1][j]=0;
       }
       ll ans=0;
       for(int i=0;i<ba[m];i++)
       {
           ans+=dp[cur][i];
           ans%=M;
       }
       printf("Case %d: %d\n",++cas,ans);
   }
    return 0;
}