hdu1561 (tree-shaped dp+01 backpack) The more, the Better

This topic just began to think for a while, because it is a tree-like DP topic, so did not see it is 01 backpack (the focus is to reverse), reverse this point has not been thought out why, because the treasure only one, and 01 backpack is the same meaning (actually I now I do not want to understand that since it is already two-dimensional, And why to reverse it), the state transfer equation is Dp[u][j]=max (dp[u][j],dp[u][j-k]+dp[v][k]), because I was in the process of building 0 as the heel, the equivalent of more than a fortress, do in order to be dp[0][m+1])
The code is as follows:

#include <iostream> #include <string.h> #include <stdio.h> using namespace std;
int n,m,edge,a;
const int maxn=205;
int DP[MAXN][MAXN];
int HEAD[MAXN],VAL[MAXN];
    struct node {int to;
int next;
}TREE[MAXN];
    void Init () {memset (dp,0,sizeof (DP));
    memset (head,-1,sizeof (head));
edge=0;
    } void Addedge (int u,int v) {tree[edge].to=v;
    Tree[edge].next=head[u];
head[u]=edge++;
    } void Dfs (int u) {dp[u][1]=val[u];
        for (int i=head[u];i!=-1;i=tree[i].next) {int v=tree[i].to;
        DFS (v);
    for (int. j=n;j>=1;j--) for (int k=1;k<j;k++) Dp[u][j]=max (Dp[u][j],dp[u][j-k]+dp[v][k]);
        }} int main () {while (scanf ("%d%d", &n,&m)!=eof) {if (n==0&&m==0) break;
        Init ();
            for (int i=1;i<=n;i++) {scanf ("%d%d", &a,&val[i]);
        Addedge (A,i);
        } dfs (0);
    printf ("%d\n", dp[0][m+1]);
}    return 0;
 }