[Bzoj1079] for coloring

Since the final staining with only ci is the number of a few, and therefore the definition of the state of f [a] [b] [ c] [d] [e] [p] expressed a a =. 1 ci , B a = 2 ci , ...... with e a CI =. 5 , the last selected ci = p a. State transfer: find p make p-1 less choice once, so we can write equation (see codes) can be used to write memory search.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 long long n,k,s[6],f[16][16][16][16][16][6];
 4 long long dfs(int a,int b,int c,int d,int e,int p){
 5     if (f[a][b][c][d][e][p])return f[a][b][c][d][e][p];
 6     if (a)f[a][b][c][d][e][p]+=(a-(p==2))*dfs(a-1,b,c,d,e,1);
 7     if (b)f[a][b][c][d][e][p]+=(b-(p==3))*dfs(a+1,b-1,c,d,e,2);
 8     if (c)f[a][b][c][d][e][p]+=(c-(p==4))*dfs(a,b+1,c-1,d,e,3);
 9     if (d)f[a][b][c][d][e][p]+=(d-(p==5))*dfs(a,b,c+1,d-1,e,4);
10     if (e)f[a][b][c][d][e][p]+=e*dfs(a,b,c,d+1,e-1,5);
11     return f[a][b][c][d][e][p]=f[a][b][c][d][e][p]%1000000007+1000000007;
12 }
13 int main(){
14     scanf("%lld",&n);
15     for(int i=1;i<=n;i++){
16         scanf("%lld",&k);
17         s[k]++;
18     }
19     for(int i=1;i<=5;i++)f[0][0][0][0][0][i]=1;
20     printf("%lld",dfs(s[1],s[2],s[3],s[4],s[5],0)-1000000007);
21 }

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