Description

Solution

樹形依賴揹包裸模型

F[i][j]表示表示DFS序上第i~N個已經做完了，重量為j
F[i][j]=max(F[i+1][j−weight[d[i]]]+value[d[i]],F[i+size[d[i]]][j])

Code

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
 

#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fod(i,a,b) for(int i=a;i>=b;i--)
#define N 2005
using namespace std;
int fs[N],dt[2*N],nt[2*N],f[N][N],n,m,m1,vl[N],pr[N],dfn[N],n1,sz[N];
void link(int x,int y)
{
    nt[++m1]=fs[x];
    dt[fs[x]=m1]=y;
}
void dfs(int k,int fa)
{
    dfn[++n1]=k;
 

    sz[k]=1;
    for(int i=fs[k];i;i=nt[i])
    {
        int p=dt[i];
        if(p!=fa) dfs(p,k),sz[k]+=sz[p];
    }
}
int main()
{
    cin>>n>>m;
    fo(i,1,n) scanf("%d%d",&vl[i],&pr[i]);
    fo(i,1,n-1)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        link(x,y),link(y,x);
 

    }
    dfs(1,0);
    fod(i,n,1)
    {
        fod(j,m,0)
        {
            f[i][j]=f[i+sz[dfn[i]]][j];
            if(j>=pr[dfn[i]]) f[i][j]=max(f[i][j],f[i+1][j-pr[dfn[i]]]+vl[dfn[i]]);
        }
    }
    int ans=0;
    fo(i,0,m) ans=max(ans,f[1][i]);
    printf("%d\n",ans);
}