"More than 2016 schools" T2 forest (tree dp, number theory)

Test instructions: There is a tree of n points, each point on a bit of right

Defines the sum of the points of the path length for all points passing through, the diameter of the tree is the largest path length in a tree

There are n inquiries, each asking for answers to the product of the diameter of all trees

One edge is deleted after each inquiry, and the number of trees is +1.

Ask to answer n times, answer mod 10^9+7

n<=100000

Idea: Because know each time to delete which side so you can do it offline, add one side at a time

The Edge will merge the two trees, considering the merging of the trees.

Knowing the shape of the original tree, the parent-child relationship between points

Consider DP, set dp[u,1],dp[u,2] to be the root of u, the longest and the minimum of the path in the subtree, and record from which son to fetch

F[u] is the maximum length of a path in a subtree that is rooted in U

Note that the longest with the second must be transferred by two different sons, which are updated attention

For ans[i] to merge x and Y, they must be parent-child relationships in the original tree

Father with X as Y

The new ans is the old ans/old f[y]* new f[the root of the subtree, that is, the topmost root node of x]

Recursion up

The division modulus is appeared, and the inverse element is obtained using the Fermat theorem

a^ (p-2) =a^-1 (mod p)

The solution is brute force + multiplication optimization, with a diameter of four points to the maximum value of U1,V1,U2,V2

But I'm too lazy to write

1 Constmo=1000000007;2 varG:Array[1..100000,1..2] ofInt64;3H:Array[1..100000,1..2] ofLongint;4FA,CX,CY,B,FF:Array[1..100000] ofLongint;5F,a,ans:Array[1..100000] ofInt64;6Head,vet,next:Array[1..300000] ofLongint;7 N,i,x,y,tot:longint;8 T:int64;9 Ten procedureAdd (a,b:longint); One begin A Inc (TOT); -next[tot]:=Head[a]; -vet[tot]:=b; thehead[a]:=tot; - End; -  - functionMax (X,y:int64): Int64; + begin -  ifX>y Thenexit (x); + exit (y); A End; at  - procedureSwapvarx,y:longint); - varT:longint; - begin -T:=x; X:=y; y:=T; - End; in  - procedureDFS (u,pre:longint); to varE,v:longint; + begin -e:=Head[u]; the   whileE<>0  Do *  begin $v:=Vet[e];Panax Notoginseng   ifV<>pre Then -   begin theff[v]:=u; + DFS (v,u); A   End; thee:=Next[e]; +  End; - End; $  $ functionmi (x,y:int64): Int64; - varTmp:int64; - begin themi:=1; tmp:=x; -   whileY>0  DoWuyi  begin the   ifY and 1=1  ThenMi:=mi*tmpMoDmo; -Tmp:=tmp*tmpMoDmo; WuY:=y>>1; -  End; About End; $  - functionexf (X:int64): Int64; - begin -Exit (MI (x,mo-2)); A End; +  the begin -Assign (input,'forest.in'); Reset (input); $Assign (output,'Forest.out'); Rewrite (output); the READLN (n); the   fori:=1  toN Do the  begin the read (a[i]); -f[i]:=A[i]; in  End; the   fori:=1  toN-1  Do the  begin About read (cx[i],cy[i]); the Add (Cx[i],cy[i]); the Add (Cy[i],cx[i]); the  End; +   fori:=1  toN-1  Doread (b[i]); -Dfs1,-1); thet:=1;Bayi   fori:=1  toN DoT:=t*a[i]MoDmo; theans[n]:=T; the   fori:=n-1 Downto 1  Do -  begin -X:=cx[b[i]]; y:=Cy[b[i]]; the   ifFf[y]<>x Thenswap (x, y); theT:=T*EXF (F[y])MoDmo; thefa[y]:=x; the    whileX>0  Do -   begin the    iffa[x]=0  ThenT:=T*EXF (F[x])MoDmo; thef[x]:=Max (f[x],f[y]); the    ifG[y,1]+a[y]>g[x,1] Then94    begin the     ifH[x,1]<>y Then the     begin theG[x,2]:=g[x,1]; H[x,2]:=h[x,1];98     End; AboutG[x,1]:=g[y,1]+A[y]; -H[x,1]:=y;101    End102     Else if(G[y,1]+a[y]>g[x,2]) and(Y<>h[x,1]) Then103     begin104G[x,2]:=g[y,1]+A[y]; theH[x,2]:=y;106     End;107F[x]:=max (F[x],g[x,1]+g[x,2]+a[x]);108Y:=x; x:=Fa[x];109   End; theT:=t*f[y]MoDmo;111ans[i]:=T; the  End;113   fori:=1  toN DoWriteln (Ans[i]); the  the close (input); the close (output);117 End.

"More than 2016 schools" T2 forest (tree dp, number theory)