[T2] 2019.8.20 NOIP simulation game B small tree (Tree) (Tree DP)

Tree \ (the DP \)

Consider setting \ (f_ {i, j, k} \) represents the \ (I \) within the sub-tree, from \ (I \) down the length of the longest chain \ (J \) , \ (I \ ) inner diameter subtree of length \ (K \) probability.

Then we will be able to find this thing a direct transfer is almost impossible.

Therefore, when we want to open the secondary transfer array \ (S_ {OP, X, Y, K} \) , where \ (OP \) for accumulated, represent the longest chain \ (X \) , long chain secondary \ (Y \) , the length of the child node in the subtree diameter less \ (K \) probability.

Then we just enumerate child nodes, then enumerate child nodes of a chain length in the sub-tree, table method can be employed simply brush \ (the DP \) transferred.

Seemingly \ (O (n-^. 5) \) , but if you remember employed (Size \) \ The method of optimizing bound on the transfer, depending on the complexity of the tree backpack, this should be \ (^ O (n 4 ) \) is.

Finally, we update \ (f_ {x, i, max (i + j, k)} \) plus \ (s_ {op, i, j, k} -s_ {op, i, j, k-1} \ ) .

Code

#include<bits/stdc++.h>
#define Tp template<typename Ty>
#define Ts template<typename Ty,typename... Ar>
#define Reg register
#define RI Reg int 
#define Con const
#define CI Con int&
#define I inline
#define W while
#define N 60
#define DB double
#define add(x,y) (e[++ee].nxt=lnk[x],e[lnk[x]=ee].to=y)
#define Gmax(x,y) (x<(y)&&(x=(y)))
using namespace std;
int n,ee,lnk[N+5];struct edge {int to,nxt;}e[N<<1];
class DpSolver
{
    private:
        int g[N+5],l[N+5];DB f[N+5][2*N+5][2*N+5],s[2][2*N+5][2*N+5][2*N+5];
        I void DP(CI x,CI lst)
        {
            RI i,j,k,p,q,op;DB t,v;
            for(g[x]=1,i=lnk[x];i;i=e[i].nxt) e[i].to^lst&&(DP(e[i].to,x),g[x]+=g[e[i].to]);//DP子节点
            for(i=0;i<=2*g[x];++i) s[0][0][0][i]=1;//初始化
            for(op=0,i=lnk[x];i;i=e[i].nxt) if(e[i].to^lst)//枚举子节点
            {
                for(j=0;j<=l[e[i].to];++j) for(k=1;k<=2*g[x];++k) f[e[i].to][j][k]+=f[e[i].to][j][k-1];//处理前缀和，方便转移
                for(op^=1,j=0;j<=l[x];++j) for(k=0;k<=j;++k) for(p=0;p<=2*g[x];++p)//枚举当前状态
                {
                    t=s[op^1][j][k][p],s[op^1][j][k][p]=0;//记下当前状态，注意清空原先数组
                    for(q=0;q<=l[e[i].to];++q) v=0.5*t*f[e[i].to][q][p],//枚举子树内最长链
                        q+1>j?(s[op][q+1][j][p]+=v):(q+1>k?s[op][j][q+1][p]+=v:s[op][j][k][p]+=v),//如果这条边长度为1
                        q+2>j?(s[op][q+2][j][p]+=v):(q+2>k?s[op][j][q+2][p]+=v:s[op][j][k][p]+=v);//如果这条边长度为2
                }Gmax(l[x],l[e[i].to]+2);
            }
            for(i=0;i<=l[x];++i) for(j=0;j<=i;++j) for(k=2*g[x];~k;--k)//枚举状态
                f[x][i][max(i+j,k)]+=s[op][i][j][k]-(k?s[op][i][j][k-1]:0),s[op][i][j][k]=0;//将状态更新到f数组中，并清空
        }
    public:
        I void Solve()
        {
            RI i,j;DB ans=0;DP(1,0);//树形DP
            for(i=0;i<=l[1];++i) for(j=i;j<=2*g[1];++j) ans+=f[1][i][j]*j;//统计答案
            printf("%.8lf",ans);//输出答案
        }
}D;
int main()
{
    freopen("tree.in","r",stdin),freopen("tree.out","w",stdout);
    RI i,x,y;for(scanf("%d",&n),i=1;i^n;++i) scanf("%d%d",&x,&y),add(x,y),add(y,x);//读入建边
    return D.Solve(),0;
}