dp[i][we] 代表節點為i 給k個消費者提供訊號的所賺錢術 這時we就是揹包容量，每個子節點的每個dp[son][j] 0 <=j<= num[j]; 就是各個物品了

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#define mini -100000000
#define maxn 3005
using namespace std;
int dp[maxn][maxn], n, m, va[maxn], num[maxn];
int last[maxn * 2
 
], Next[maxn * 2], edge[maxn * 2], cost[maxn * 2], l;
void add(int a, int b, int va)
{
     edge[l] = b;
     cost[l] = va;
     Next[l] = last[a];
     last[a] = l;
     l++;
}
void dfs(int pre)
{
     for(int i = last[pre]; i != -1; i = Next[i])
     {
         dfs(edge[i]);
         num[pre] += num[edge[i]];
         for
 
(int j = num[pre]; j >= 0; j--)
             for(int k = min(j ,num[edge[i]]); k >= 0; k--)
                 dp[pre][j] = max(dp[pre][j], dp[edge[i]][k] + dp[pre][j - k] - cost[i]);
     }
}
int main()
{
    while(scanf("%d%d", &n, &m)!=EOF)
    {
        l = 0;
        memset(last, -1, sizeof
 
(last));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= n; j++) dp[i][j] = mini;
        for(int i = 1; i<= n - m; i++)
        {
            num[i] = 0;
            int k;
            scanf("%d", &k);
            while(k--)
            {
               int a, b, c;
               scanf("%d%d", &a, &b);
               add(i, a, b);
            }
        }
        for(int i = n - m + 1; i <= n; i++)
        {
            num[i] = 1;
            scanf("%d", &dp[i][1]);
        }
        dfs(1);
        for(int i = m; i >= 0; i--) if(dp[1][i] >= 0) {printf("%d\n", i); break;}
    }
    return 0;
}