POJ 2486 Apple trees (tree dp, tree backpack)

Test instructions: Given a tree map, a person from point s, can only walk the K-step, each point has a certain number of apples, asked to collect as many apples as possible, output the maximum number of apples.

Ideas:

Since it is a tree, and there are restrictions on the K-step, then the tree-shaped DP just.

Consider a situation where: (1) may be after the completion of Step K (2) at the end of the tree, and then go back to this node (step k is not in the book tree)

The second kind of simple, backpack, is to enumerate to this node's children t How many steps, collect up to the maximum number of apples. The first requires that step K be terminated in one of the sub-trees under this node, then only in 1 children's subtree, so it should be "other children all have to come back" + "the child does not ask to come back" or "some other child does not walk back" + "This node comes back."

Use two DP arrays to differentiate between "back" and "not back" on the line, note, "Do not return" only 1 children do not require it to walk back, "back" is all back. The number of apples collected "without asking for a return" must be greater than or equal to "request back".

1 //#include <bits/stdc++.h>2#include <cstdio>3#include <cstring>4#include <iostream>5 #definePII pair<int,int>6 #defineINF 0x3f3f3f3f3f3f3f3f7 #defineLL Long Long8 using namespacestd;9 Const intn= About;Ten structnode One { A     int  from, To,next; - node () {}; -Nodeint  from,intTo,intNext): from( from), to, next (next) {}; the}edge[n*2]; - intedge_cnt, Head[n], w[n]; - voidAdd_node (int  from,intto ) - { +Edge[edge_cnt]=node ( from, To, head[ from]); -head[ from]=edge_cnt++; + } A /* at Dp[][][1] Records are returned to this node each time.  - Dp[][][0] records only 1 times without returning to this node.  - */ - intdp[n][n][2]; - voidDFS (intTintFarintm) - { in node E; -      for(intj=0; j<=m; J + +) dp[t][j][0]=dp[t][j][1]=W[T];//Now that we can get here, at least take this node. to     if(m==0)return; +      for(intI=HEAD[T]; i!=-1; I=e.next) -     { theE=Edge[i]; *         if(e.to^Far ) $         {Panax NotoginsengDFS (e.to, T, M-1); -              for(intJ=m; J>0; j-- ) the             { +                  for(intk=0; K +2<=j; k++ ) A                 { the                     //all branches back.  +dp[t][j][1]=max (dp[t][j][1], dp[t][j-k-2][1]+dp[e.to][k][1] ); -                     //This branch is going back, but not back in the other branches. Because there are already 1 not back, so update in ' [0] '.  $dp[t][j][0]=max (dp[t][j][0], dp[t][j-k-2][0]+dp[e.to][k][1] ); $                 } -                  for(intk=0; K +1<=j; k++ ) -                 { the                     //not back, but the other branches must be all back.  -dp[t][j][0]=max (dp[t][j][0], dp[t][j-k-1][1]+dp[e.to][k][0] );Wuyi                 } the             } -         } Wu     } - } About  $  -  - intMain () - { A     //freopen ("Input.txt", "R", stdin); +     intN, K, A, b; the      while(~SCANF ("%d%d",&n,&K)) -     { $Edge_cnt=0; theMemset (Head,-1,sizeof(head)); the  the          for(intI=1; i<=n; i++) scanf ("%d",&w[i]); the          for(intI=1; i<n; i++) -         { inscanf"%d%d",&a,&b); the Add_node (A, b); the Add_node (b,a); About         } theDFS (1, -1, K); thecout<<dp[1][k][0]<<Endl; the     } +     return 0; -}

AC Code

POJ 2486 Apple trees (tree dp, tree backpack)