Apple Tree POJ-2486

Apple Tree POJ-2486

The main topic: a point with the right to have a root tree, the root node is 1. From the root node, take the K-step, to collect the maximum weight and.

Tree-shaped DP. The complexity may be O (metaphysics) and will not exceed $o (nk^2) $. (Anyway, this question does not card this, the examination thought) reference

Ans[i][j][0] means I point below a total of walk J step, do not come back, may collect the maximum weight value
Ans[i][j][1] means I point below a total of walk J step, come back, may collect the maximum weight value

It is more complicated to use a backpack when each node (hereinafter referred to as the current node) is transferred from its child nodes:

T[i][j][0] Indicates that the front I child node of the current node is walking J step, not coming back
T[I][J][1] Indicates that the front I sub-node of the current node is walking J step back

For T[i][j][0], either the former I-1 sub-node of the current node takes J-step (including the number of steps to go and back to the child nodes in front of it), and does not come back before;

Either the former I-1 sub-node j-p step (including the number of steps to go and back to the front of the child node), the current node to go to the sub-node I 1 steps, I sub-node p-1 step down, do not come back;

Either take a step to the sub-node I, the sub-node I walk down p-2 step, then take a step back to the current node, and then in the former I-1 sub-node j-p step (including the number of steps to go and back to the front of the child node) and do not come back.

So T[i][j][0]=max (T[i-1][j][0],max{t[i-1][j-p][1]+ans[nowson][p-1][0]},max{t[i-1][j-p][0]+ans[nowson][p-2][1]})

For T[i][j][1], either the former I-1 sub-node j-p step (including the number of steps to go and back to the front of the child node), walk to the sub-node I took 1 steps, I sub-node walk down with P-2 step and back, from the sub-node I back to take a step Or the former I-1 sub-node takes a J-step (including the number of steps to go and back in front of the child nodes), back.

So T[i][j][1]=max (T[i-1][j][1],max{t[i-1][j-p][1]+ans[nowson][p-2][1]})

Of course, the actual solution does not require each node to open a T array, just need to do on the ANS array directly on the line. Is the first to solve the T array process with a scrolling array optimization, then only two D T[J][0/1]. Just take the array of ans[current node] as T to do it. Also, consider the bounds of the T array. In addition, the T array is initialized to the current node weight before it is solved.

The final answer is obvious: Max (Ans[1][k][0],ans[1][k][1]).

Once error:

Transfer equation of naive:

T[i][j][0]=max (T[i-1][j][0],t[i-1][j-p][0],t[i-1][j-p][1]+ans[son][p][0])
T[i][j][1]=max (T[i-1][j][1],t[i-1][j-p][1]+ans[son][p][1])

In fact, the 3rd (red) Case of the transfer of t[i][j][0] is easy to miss. In addition, it is easy to ignore the 1 or 2 steps that are spent walking and walking back to the child nodes.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 structEdge6 {7     intTo,next;8}edge[ About];9 intne,ans[ the][ About][2],f1[ the];Ten inta[ the]; One intn,k; A BOOLvis[ the]; - voidDfsintu) - { the     intj,kk=f1[u],p,v; -vis[u]=true; -      for(j=0; j<=k;j++) -ans[u][j][0]=ans[u][j][1]=A[u]; +      while(kk!=0) -     { +v=edge[kk].to; A         if(!Vis[v]) at         { - Dfs (v); -              for(j=k;j>=0; j--) -             { -                  for(p=1;p <=j;p++) -ans[u][j][0]=max (ans[u][j][0],max (ans[u][j-p][0]+ans[v][p-2][1],ans[u][j-p][1]+ans[v][p-1][0])); in                  for(p=2;p <=j;p++) -ans[u][j][1]=max (ans[u][j][1],ans[u][j-p][1]+ans[v][p-2][1]); to             } +         } -kk=Edge[kk].next; the     } * } $ intMain ()Panax Notoginseng { -     intI,TA,TB; the      while(SCANF ("%d%d", &n,&k) = =2) +     { ANe=0; thememset (ans,0,sizeof(ans)); +memset (Vis,0,sizeof(Vis)); -memset (F1,0,sizeof(F1)); $          for(i=1; i<=n;i++) $scanf"%d",&a[i]); -          for(i=1; i<n;i++) -         { thescanf"%d%d",&ta,&TB); -edge[++ne].to=TB;Wuyiedge[ne].next=F1[ta]; thef1[ta]=NE; -edge[++ne].to=ta; Wuedge[ne].next=F1[TB]; -f1[tb]=NE; About         } $Dfs1); -printf"%d\n", Max (ans[1][k][0],ans[1][k][1])); -     } -     return 0; A}

Apple Tree POJ-2486