POJ2486 Apple Tree "backpack"

A word test instructions: a tree, a total of n points, each point has a weight, ask to start from 1, walk the K-step, up to the right value to traverse. can go back.

The first (second) road tree knapsack problem, first looked at the Dalao, changed a bit on the example. However.... TLE??? It's been a while. found that because of multiple sets of data and no "0 0" input, if not read in the time to add "~" or "EOF" will be dead loop, resulting in tle.

State design: Set F[I][J][0/1] As the root of the subtree, Walk J, can get the maximum weight (0/1 of the expression will be described in the transfer equation) consider: (here refer to [email protected], invasion and deletion of QWQ) each node in the final answer type: 1, do not pass; 2, passing but not returning; 3, passing and returning (the definition returned is that the final stop node is not in the subtree of that node) then it can be transferred:

Then there are some details: assigning an initial value to an array (starting with 0!) Walk 0 steps) to update the head array.

The implementation is still relatively simple.

Code

1#include <cstdio>2#include <algorithm>3#include <cstring>4 5 using namespacestd;6 7 intN,k,tot;8 intw[ -],head[ -];9 intf[ -][ $][3]; Ten structnode{ One     intTo,next,val; A}edge[ -]; -  - voidAddintXinty) the { -edge[++tot].to=y; -edge[tot].next=Head[x]; -head[x]=tot; + } -  + voidReadint&x) A { atx=0; -     CharCh=GetChar (); -     BOOLflag=false; -      while(ch<'0'|| Ch>'9') flag|= (ch=='-'), ch=GetChar (); -      while(ch>='0'&&ch<='9') x= (x<<3) + (x<<1) + (ch^ -), ch=GetChar (); -X=flag? -x:x; in } -  to voidTREEDP (intUintFA) + { -      for(intI=0; i<=k;i++) f[u][i][1]=w[u],f[u][i][0]=W[u]; the      for(intI=head[u];i;i=edge[i].next) *     { $         intv=edge[i].to;Panax Notoginseng         if(V==FA)Continue; - TREEDP (v,u); the          for(intkk=k;kk>=0; kk--) +          for(intj=0; j<=kk;j++) A         { the             if(kk>=j+2) f[u][kk][0]=max (f[u][kk][0],f[v][j][0]+f[u][kk-j-2][0]); +             if(kk>=j+2) f[u][kk][1]=max (f[u][kk][1],f[v][j][0]+f[u][kk-j-2][1]); -             if(kk>=j+1) f[u][kk][1]=max (f[u][kk][1],f[v][j][1]+f[u][kk-j-1][0]); $         } $     } - } -  the voidInit () - {WuyiMemset (F,0,sizeof(f)); theMemset (Head,0,sizeof(head)); -tot=0; Wu }  -  About intMain () $ { -      while(SCANF ("%d%d", &n,&k)! =EOF) -     { -          for(intI=1; i<=n;i++) read (W[i]); A          for(intI=1; i<=n-1; i++) +         { the             intx=0, y=0; - read (x), read (y); $ Add (x, y), add (y,x); the         } theTREEDP (1,-1); theprintf"%d\n", Max (f[1][k][0],f[1][k][1])); the init (); -     } in     return 0; the}

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Summary: The classification discussion is often also an important breakthrough in solving the problem.

POJ2486 Apple Tree "backpack"