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-8
# define MOD 1000000007
# define INF 1000000000
# 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<<1,l,mid
# define rch p<<1|1,mid+1,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=0, flag=0;
    char ch;
    if((ch=getchar())==‘-‘) flag=1;
    else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘;
    while((ch=getchar())>=‘0‘&&ch<=‘9‘)  res=res*10+(ch-‘0‘);
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar(‘-‘); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+‘0‘);
}
const int N=20;
//Code begin...

int vis[6];
LL dp[16][16][16][16][16][6];
bool mark[16][16][16][16][16][6];

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=0;
    if (a) ans+=dfs(a-1,b,c,d,e,1)*(last==2?a-1:a);
    if (b) ans+=dfs(a+1,b-1,c,d,e,2)*(last==3?b-1:b);
    if (c) ans+=dfs(a,b+1,c-1,d,e,3)*(last==4?c-1:c);
    if (d) ans+=dfs(a,b,c+1,d-1,e,4)*(last==5?d-1:d);
    if (e) ans+=dfs(a,b,c,d+1,e-1,5)*e;
    mark[a][b][c][d][e][last]=1;
    return dp[a][b][c][d][e][last]=ans%MOD;
}
int main ()
{
    int n, x;
    scanf("%d",&n);
    FOR(i,1,n) scanf("%d",&x), ++vis[x];
    FOR(i,1,5) dp[0][0][0][0][0][i]=1, mark[0][0][0][0][0][i]=1;
    dfs(vis[1],vis[2],vis[3],vis[4],vis[5],0);
    printf("%lld\n",dp[vis[1]][vis[2]][vis[3]][vis[4]][vis[5]][0]);
    return 0;
}