Farmer and money

There are N number of farmers who live in N different villages. This N villages form a tree to get the money when the initial X each farmer. Each one operation, a farmer from its own money in, take out any amount of money handed over to a neighboring village farmers. for each farmer given a value v_i, seeking (1) requires a minimum number of operations, so that each farmer finally got the money> = a given value v_i .

Fool knows is a tree dp, but have to work around it

F [i] [j] denotes the subtree i, j with the operations, in that each farmer i eventually got money> = a given value of v conditions give father how much

Backpack tree do: f '[i] [j + k] = f [i] [k] + f [v] [j]

Get

#include<bits/stdc++.h>
using namespace std;
int n;
long long x,f[2001][2001],val[2001],g[2001];
int sz[2001]; 
vector<int> e[2001];
void merge(int x,int y){
    int i,j; 
    for(i=0;i<sz[x];i++)g[i]=f[x][i];
    for(i=0;i<sz[x]+sz[y];i++)f[x][i]=-2147483648ll;
    for(i=0;i<sz[x];i++){
        for(j=0;j<sz[y];j++){
            if(f[y][j]>=0)f[x][i+j]=max(f[x][i+j],g[i]);
            f[x][i+j+1]=max(f[x][i+j+1],g[i]+f[y][j]);
        } 
    }sz[x]+=sz[y];
    for(i=1;i<sz[x];i++)f[x][i]=max(f[x][i-1],f[x][i]);
}
void dfs(int u,int fa){
    int i,j;
    f[u][0]=f[u][1]=x-val[u],sz[u]=1;
    int v,k;
    for(i=0;i<e[u].size();i++){
        v=e[u][i];
        if(v==fa)continue;
        dfs(v,u),merge(u,v);
    }
} 
int main(){
    int i;
    cin>>n>>x;
    for(i=1;i<=n;i++)scanf("%lld",&val[i]);
    for(i=1;i<n;i++){
        int a,b;
        scanf("%d%d",&a,&b);
        e[a].push_back(b),e[b].push_back(a);    
    }
    dfs(1,0);
    for(i=0;i<n;i++)if(f[1][i]>=0)break;
    cout<<i<<endl;
    return 0;
}