POJ 1155 TELE Backpack type Tree DP Classic

The system of television, transit, and user's TV is just a tree.

n nodes, numbered respectively 1~n,1 is the TV station Center, 2~n-m is the broker, N-m+1~n is the user, 1 is the root

Now node 1 ready to broadcast a game, known to transfer data from one node to another node, the TV station needs a certain cost

If the data can be transferred to the user's node n-m+1~n, these users are willing to pay a certain fee to the TV station

Now the TV station wants to broadcast the competition for as many users as possible without losing money.

Output up to how many users broadcast the game

Tree DP first question of backpack type

DP[I][J] indicates that a subtree with node i is the root of a tree with J users receiving retransmission, the maximum benefit of the TV station

Since there is a positive negative income

Initialization

Dp[i][0]=0

Dp[i][j]=-inf (j>0)

Target: Dp[1][j]>=0 under the condition of the largest J

(Dp[1][j]>=0 said the TV station does not lose money)

Process of Dfs recursive DP

There's an idea where the notes are written.

1#include <cstdio>2#include <cstring>3 4 using namespacestd;5 6InlineintMaxintAintb)7 {8     returnA>b?a:b;9 }Ten  One Const intmaxn=3005; A Const intinf=0x3f3f3f3f; -  - intDP[MAXN][MAXN]; the intCOST[MAXN][MAXN]; - intSIZ[MAXN]; - structEdge - { +     intTo,next; - }; + Edge EDGE[MAXN]; A intHEAD[MAXN]; at inttot; - intn,m; -  - voidAddedge (intUintv) - { -edge[tot].to=v; inedge[tot].next=Head[u]; -head[u]=tot++; to } +  - voidInit () the { *memset (head,-1,sizeofhead); $tot=0;Panax Notoginseng      for(intI=1; i<=n;i++) -     { thedp[i][0]=0; +          for(intj=1; j<=m;j++) Adp[i][j]=-inf; the     } + } -  $ voidsolve (); $ voidDfsintu); -  - intMain () the { -      while(~SCANF ("%d",&N))Wuyi     { thescanf"%d",&m); - init (); Wu          for(intI=1; i<=n-m;i++) -         { About             intnum; $scanf"%d",&num); -              while(num--) -             { -                 intu; Ascanf"%d",&u); +scanf"%d",&Cost[i][u]); the Addedge (i,u); -             } $         } the          for(inti=n-m+1; i<=n;i++) the         { thescanf"%d", &cost[i][maxn-1]); the         } - solve (); in     } the     return 0; the } About  the voidSolve () the { theDfs1); +      for(intj=siz[1];j>=0; j--) -     { the         if(dp[1][j]>=0)Bayi         { theprintf"%d\n", j); the             return ; -         } -     } the     /* the for (int i=0;i<=siz[1];i++) the printf ("%d\n", Dp[1][i]); the     */ - } the  the voidDfsintu) the {94siz[u]=0; the      for(intI=head[u];~i;i=edge[i].next) the     { the         intv=edge[i].to;98         if(n-m<v) About         { -dp[v][0]=0;101dp[v][1]=cost[v][maxn-1];102siz[v]=1;103         }104         Else the         {106 Dfs (v);107         }108siz[u]+=Siz[v];109          the         //due to update dp[u][j]111         //Dp[u][j-k] needs to represent a subtree with the root of U, and j-k is taken from the front son node . the         //so this time dp[u][j-k] can not be updated by the Son node V (to ensure that j-k are taken from the front)113         //So, when the recursion is pushed, J pushes the inverse . the         //Of course, the order of the K and J loops cannot be exchanged. the          for(intJ=siz[u];j>0; j--) the         {117              for(intk=1; k<=siz[v];k++)118                 if(j>=k)119Dp[u][j]=max (dp[u][j],dp[u][j-k]+dp[v][k]-cost[u][v]); -         }121     }122}

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POJ 1155 TELE Backpack type Tree DP Classic