BZOJ 1079 着色方案(DP)

如果把当前格子涂什么颜色当做转移的话，状态则是每个格子的颜色数还剩多少，以及上一步用了什么颜色，这样的状态量显然是5^15.不可取。

如果把当前格子涂颜色数还剩几个的颜色作为转移的话，状态则是每个格子的还剩多少个的颜色数，以及上一步用了还剩几个的颜色，这样的状态量为15^5.

那么定义dp[a][b][c][d][e][last].表示涂到当前格子时还剩1个的颜色数为a.......且上一步涂了还剩last个的颜色。

转移方程就很明显了.

/**************************************************************
    Problem: 1079
    User: freeloop
    Language: C++
    Result: Accepted
    Time:52 ms
    Memory:56588 kb
****************************************************************/

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-
# define MOD
# define INF
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<,l,mid
# define rch p<<|,mid+,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=, flag=;
    char ch;
    if((ch=getchar())=='-') flag=;
    else if(ch>=''&&ch<='') res=ch-'';
    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');
    return flag?-res:res;
}
void Out(int a) {
    if(a<) {putchar('-'); a=-a;}
    if(a>=) Out(a/);
    putchar(a%+'');
}
const int N=;
//Code begin...

int vis[];
LL dp[][][][][][];
bool mark[][][][][][];

LL dfs(int a, int b, int c, int d, int e, int last)
{
    if (mark[a][b][c][d][e][last]) return dp[a][b][c][d][e][last];
    LL ans=;
    if (a) ans+=dfs(a-,b,c,d,e,)*(last==?a-:a);
    if (b) ans+=dfs(a+,b-,c,d,e,)*(last==?b-:b);
    if (c) ans+=dfs(a,b+,c-,d,e,)*(last==?c-:c);
    if (d) ans+=dfs(a,b,c+,d-,e,)*(last==?d-:d);
    if (e) ans+=dfs(a,b,c,d+,e-,)*e;
    mark[a][b][c][d][e][last]=;
    return dp[a][b][c][d][e][last]=ans%MOD;
}
int main ()
{
    int n, x;
    scanf("%d",&n);
    FOR(i,,n) scanf("%d",&x), ++vis[x];
    FOR(i,,) dp[][][][][][i]=, mark[][][][][][i]=;
    dfs(vis[],vis[],vis[],vis[],vis[],);
    printf("%lld\n",dp[vis[]][vis[]][vis[]][vis[]][vis[]][]);
    return ;
}