【SCOI2008】着色方案

题目：

http://oj.changjun.com.cn/problem/detail/pid/2027

考虑记忆化搜索。

因为每种颜色能涂的木块<=5,设f[a][b][c][d][e][last]代表当前还剩ａ个能涂１的木块......上一个涂的木块是剩ｌａｓｔ的木块

则f[a,b,c,d,e,last]=(a-(last==2))*f[a-1,b,c,d,e]+(b-(last==3))*f[a+1,b-1,c,d,e]+
(c-(last==4))*f[a,b+1,c-1,d,e]+...+( e(last==6))*f[a,b,c,d+1,e-1]。

 #include<set>
 #include<map>
 #include<queue>
 #include<stack>
 #include<ctime>
 #include<cmath>
 #include<string>
 #include<vector>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define LL long long
 #define mod 1000000007
 using namespace std;
 int g[],n=,k;
 LL f[][][][][][];//第六维判重
 bool bj[][][][][][];
 LL dfs(int a,int b,int c,int d,int e,int Last)
 {
   if(a+b+c+d+e==) return ;
   if(bj[a][b][c][d][e][Last]) return f[a][b][c][d][e][Last]%mod;
   LL ans=;
   if(a) ans+=(a-(Last==))*dfs(a-,b,c,d,e,),ans%=mod;//放a
   if(b) ans+=(b-(Last==))*dfs(a+,b-,c,d,e,),ans%=mod;//放b
   if(c) ans+=(c-(Last==))*dfs(a,b+,c-,d,e,),ans%=mod;//放c
   if(d) ans+=(d-(Last==))*dfs(a,b,c+,d-,e,),ans%=mod;//放d
   if(e) ans+=(e-(Last==))*dfs(a,b,c,d+,e-,),ans%=mod;//放e
   f[a][b][c][d][e][Last]=ans%mod;
   bj[a][b][c][d][e][Last]=;
   return ans%mod;
 }
 int main()
 {
   freopen("!.in","r",stdin);
   freopen("!.out","w",stdout);
   scanf("%d",&k);
   int op;
   for(int i=;i<=k;i++)
     scanf("%d",&op),g[op]++;//能涂ｏｐ个的数量
   dfs(g[],g[],g[],g[],g[],);
   printf("%lld",f[g[]][g[]][g[]][g[]][g[]][]);
   return ;
 }