[树形依赖多重背包] BZOJ 4910 [Sdoi2017] 苹果树

首先解决依赖背包
如果是0/1背包，按照后序遍历dp，根据选或不选决策
现在是多重背包，那么这个点只留一个，剩下的变成一个新点挂上去，这样仍然满足依赖关系，
转移的时候多重背包用单调对列优化

可以发现如果除了最长的一条链，剩余最多K个
因为权值为正 那么最长链必然连到叶子，剩余必然取K个（除非不足K个）
那么我们枚举叶子 然后两边总共取出K个，枚举两边分别选几个
两边需要两次儿子顺序不同的后序遍历
复杂度O(nK)
被卡常，我把部分分特判了才过TAT

#include<cstdio>#include<cstdlib>#include<algorithm>#include<vector>#include<cstring>#define cl(x) memset(x,0,sizeof(x))#define pb push_backusing namespace std;typedef long long ll;inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(),b=1;  for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;  for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;}const int N=40005;const int KK=500005;const int NK=51000000;struct edge{  int u,v,next;}G[N<<1];int head[N],inum;inline void add(int u,int v,int p){  G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;}int a[N],v[N];#define V G[p].vint n,K;int ncnt,pos[N];int lst1[N],lst2[N],pnt;int pre1[N],pre2[N],back1[N],back2[N];int size[N];int f1[NK],f2[NK];int leaf[N];int val[N],depth[N];inline void dfs1(int u){  pre1[u]=pnt+1; size[u]=1;  for (int p=head[u];p;p=G[p].next){    val[V]=val[u]+v[V],depth[V]=depth[u]+1;    dfs1(V);    size[u]+=size[V];  }  lst1[++pnt]=u; back1[u]=pnt;}inline void dfs2(int u){  pre2[u]=pnt+1;  vector<int> v; for (int p=head[u];p;p=G[p].next) v.pb(V);  for (int i=v.size()-1;~i;i--)    dfs2(v[i]);  lst2[++pnt]=u; back2[u]=pnt;}int Q[KK],l,r;int Q2[KK];#define F(i,j) (ff[(i)*(K+1)+(j)])inline int max(int a,int b){  return a>b?a:b;}inline void DP(int *ff,int *lst,int *pre){  F(0,0)=0; for (int k=1;k<=K;k++) F(0,k)=-1<<29;  for (int i=1;i<=ncnt;i++){    int x=lst[i]; int *f=ff+i*(K+1),*g=ff+(i-1)*(K+1),*t=ff+(pre[x]-1)*(K+1);    //for (int k=0;k<=K;k++) f[k]=t[k];    memcpy(f,t,sizeof(int)*(K+1));    if (a[x]==0) continue;    if (a[x]==1){      for (int k=1;k<=K;k++)    f[k]=max(f[k],g[k-1]+v[x]);    }else{      l=r=-1;      Q[++r]=0; Q2[r]=g[Q[r]]-Q[r]*v[x];      for (int j=1;j<=K;j++){    while (l<r && Q[l+1]<j-a[x])      l++;    if (l<r)      f[j]=max(f[j],Q2[l+1]+j*v[x]);    int tmp=g[j]-j*v[x];    while (l<r && tmp>=Q2[r])      r--;    Q[++r]=j; Q2[r]=tmp;      }    }  }}namespace Work{  inline bool cmp(int x,int y){    return v[x]>v[y];  }  int idx[N];  inline void Solve(){    for (int i=1;i<=n;i++) idx[i]=i,a[i]+=a[pos[i]];    a[1]--;    int x=K+1,ans=v[1];    sort(idx+1,idx+n+1,cmp);    for (int i=1;i<=n;i++){      int t=min(x,a[idx[i]]);      x-=t; ans+=v[idx[i]]*t;      if (!x) break;    }    printf("%d\n",ans);  }}int main(){  int T;  int x,y,z;  freopen("t.in","r",stdin);  freopen("t.out","w",stdout);  read(T);  while (T--){    read(n); read(K);    ll tot=0; ncnt=n;    int maxv=0;    for (int i=1;i<=n;i++){      read(x); read(y); read(z); tot+=y;      maxv=max(maxv,x);      if (x) add(x,i,++inum),leaf[x]=1;      v[i]=z; a[i]=1;      if (y>1) pos[i]=++ncnt,v[ncnt]=z,a[ncnt]=y-1,add(i,ncnt,++inum);    }    if (maxv<=1){      Work::Solve();      cl(head); inum=0; cl(leaf); cl(pos);      continue;    }    pnt=0; val[1]=v[1]; depth[1]=1; dfs1(1);    DP(f1,lst1,pre1);    pnt=0; dfs2(1);    DP(f2,lst2,pre2);    int ans=-1<<30;    for (int i=1;i<=n;i++)      if (!leaf[i]){    int *f=f1+(K+1)*(pre1[i]+size[i]-2);    int *g=f2+(K+1)*(pre2[i]-1);    int x=min(tot-depth[i],(ll)K);    for (int k=0;k<=x;k++)      ans=max(ans,val[i]+f[k]+g[x-k]);      }    printf("%d\n",ans);    cl(head); inum=0; cl(leaf); cl(pos);  }  return 0;}