【jzoj5344】【NOIP2017模拟9.3A组】【摘果子】【树型依赖背包】

description

solution

直接树型依赖背包没什么好说的。

code

#include
#include
#include
#include
#define LF double
#define LL long long
#define ULL unsigned int
#define fo(i,j,k) for(int i=j;i<=k;i++)
#define fd(i,j,k) for(int i=j;i>=k;i--)
#define fr(i,j) for(int i=begin[j];i;i=next[i])
using namespace std;
int const mn=2000+2,mm=4000+2,inf=1e9+7;
int n,m,v[mn],p[mn],gra,begin[mn],to[mm],next[mm],size[mn],a[mn],f[mn][mn];
void insert(int u,int v){
    to[++gra]=v;
    next[gra]=begin[u];
    begin[u]=gra;
}
void dfs(int p,int q){
    a[++a[0]]=p;
    size[p]=1;
    fr(i,p)if(to[i]!=q){
        dfs(to[i],p);
        size[p]+=size[to[i]];
    }
}
int main(){
    freopen("d.in","r",stdin);
    freopen("d.out","w",stdout);
    scanf("%d%d",&n,&m);
    fo(i,1,n)scanf("%d%d",&v[i],&p[i]);
    fo(i,1,n-1){
        int u,v;
        scanf("%d%d",&u,&v);
        insert(u,v);
        insert(v,u);
    }
    dfs(1,0);
    fo(i,0,n+1)fo(j,0,m)f[i][j]=-inf;
    f[1][0]=0;
    fo(i,1,n){
        int q=a[i];
        fo(j,0,m){
            if(j+p[q]<=m){
                f[i+1][j+p[q]]=max(f[i+1][j+p[q]],f[i][j]+v[q]);
                f[i+size[q]][j+p[q]]=max(f[i+size[q]][j+p[q]],f[i][j]+v[q]);    
            }
            f[i+size[q]][j]=max(f[i+size[q]][j],f[i][j]);
        }
    }
    int ans=0;
    fo(j,0,m)ans=max(ans,f[n+1][j]);
    printf("%d",ans);
    return 0;
}