【WIkiOI】【P2926】【黑白瓷砖】【Polya定理】【题解】

传送门：http://www.wikioi.com/problem/2926/

Polya定理，看到高精度就不想写C++……python打了个表……，题解&C++源码有空再发

import pdb
for n in range(1,21):
    v=[[0]*22 for i in range(22)]
    tmp=[[0]*22 for i in range(22)]
    Hash=[[0]*22 for i in range(22)]
    a=[0]*444
    vis=[0]*444
    tot=1
    ans=0
    global res
    res=0
    def turn():
        for i in range(1,n+1):
            for j in range(1,n+1):
                tmp[i][j]=v[i][j]
        for i in range(1,n+1):
            for j in range(1,n+1):
                v[i][j]=tmp[n-j+1][i-j+1]
    def turnover():
        for i in range(1,n+1):
            for j in range(1,n+1):
                tmp[i][j]=v[i][j]
        for i in range(1,n+1):
            for j in range(1,n+1):
                v[i][j]=tmp[i][i-j+1]
      
    def dfs(x):
        while vis[x]==0:
            vis[x]=1
            x=a[x]
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1
    #Case 1
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]

    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1  
    #pdb.set_trace()
    #Case 2
    #pdb.set_trace()
    turn()
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]

    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1       
    #Case 3
    turn()
    turn()
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]
    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1         
    #Case 4
    turnover()
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]
    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1          
    #Case 5
    turn()
    turnover()
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]
    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1          
    #Case 6
    turn()
    turn()
    turnover()
    for i in range(1,n+1):
        for j in range(1,i+1):
            a[Hash[i][j]]=v[i][j]
    for i in range(1,tot):
        if vis[i]==0:
            res+=1
            dfs(i)
    ans+=2**res
    res=0
    vis=[0]*444
    tot=1
    for i in range(1,n+1):
        for j in range(1,i+1):
            Hash[i][j]=tot
            v[i][j]=tot
            tot+=1  
    print 'f['+str(n)+']="'+str(ans/6)+'";'