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

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description

solution

直接树型依赖背包没什么好说的。

code

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#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;}

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