Binary Apple Tree_解題報告_URAL1018_樹狀dp

標籤：

Description

Let‘s imagine how apple tree looks inbinary computer world. You‘re right, it looks just like a binary tree, i.e. anybiparous branch splits up to exactly two new branches. We will enumerate byintegers the root of binary apple tree, points of branching and the ends oftwigs. This way we may distinguish different branches by their ending points.We will assume that root of tree always is numbered by 1 and all numbers usedfor enumerating are numbered in range from 1 to N, where N is the total numberof all enumerated points. For instance in the picture below N is equal to 5.Here is an example of an enumerated tree with four branches:

2   5

 \ /

 3   4

   \/

    1

As you may know it‘s not convenient to pickan apples from a tree when there are too much of branches. That‘s why some ofthem should be removed from a tree. But you are interested in removing branchesin the way of minimal loss of apples. So your are given amounts of apples on abranches and amount of branches that should be preserved. Your task is todetermine how many apples can remain on a tree after removing of excessivebranches.

Input

First line of input contains two numbers: Nand Q (2 ≤ N ≤ 100; 1 ≤ Q ≤ N ? 1). N denotes the number of enumerated points in a tree. Qdenotes amount of branches that should be preserved. Next N ? 1 lines containsdescriptions of branches. Each description consists of a three integer numbersdivided by spaces. The first two of them define branch by it‘s ending points.The third number defines the number of apples on this branch. You may assumethat no branch contains more than 30000 apples.

Output

Output should contain the only number —amount of apples that can be preserved. And don‘t forget to preserve tree‘sroot ;-)

SampleInput

5 2

1 3 1

1 4 10

2 3 20

3 5 20

SampleOutput

21

 

題目大意：

有一顆擁有n關係的帶權二叉樹要剪去q根枝條求整棵樹的最大權值和。

 

思路：

因為是求最優值所以想到dp，因為是赤裸裸的二叉樹所以想到赤裸裸的樹狀dp，因為是赤裸裸的樹狀dp所以放上赤裸裸的狀態轉移方程。

f[i,j]:=max(f[i.left,k]+a[I,i.left],f[i.right,j-k]+a[I,i.right]),0<=k<=j

其中f[I,j]表示以i為根節點，剪去j根枝條的最優值，a[I,j]表示串連I、j枝條上的權值。F[1,q]即所求答案。注意要在開始轉換成一顆二叉樹。

 

原始碼/pas

 

typetree=record  l,r:longint;end;var  n,q:longint;  a,f,v:array[0..100,0..100]of longint;  t:array[0..100]of tree;function max(x,y:longint):longint;begin  if x>y then  max:=x  else  max:=y;end;procedure init;var  i,x,y,z:longint;begin  readln(n,q);  for i:=1 to n-1 do  begin    readln(x,y,z);    a[x,y]:=z;    a[y,x]:=z;    v[x,y]:=1;    v[y,x]:=1;  end;end;procedure create(x:longint);var  i:longint;begin  for i:=1 to n do  if (v[x,i]<>0) then  begin    if (t[x].l=0) then t[x].l:=i    else    if (t[x].r=0) then t[x].r:=i;    v[i,x]:=0;    v[x,i]:=0;    create(i);  end;end;procedure dp;var  i,j,k:Longint;begin  for i:=1 to n do  f[i,1]:=max(a[i,t[i].l],a[i,t[i].r]);  for j:=2 to q do  for i:=1 to n do  begin    f[i,j]:=max(a[i,t[i].r]+f[t[i].r,j-1],a[i,t[i].l]+f[t[i].l,j-1]);    for k:=0 to j-2 do    f[i,j]:=max(f[i,j],f[t[i].l,k]+f[t[i].r,j-2-k]+a[i,t[i].l]+a[i,t[i].r]);  end;end;begin  init;  create(1);  dp;  writeln(f[1,q]);end.

Binary Apple Tree_解題報告_URAL1018_樹狀dp