codeforces 293E 树上点分治+bit维护二维偏序

https://codeforc.es/contest/293/problem/E

题解:

如果我们和poj1741一样,暴力预处理出子树中所有的点对信息

那么问题就变成了一个二维偏序问题

多少个点对满足w1+w2<=W且l1+l2<=L

参考树状数组求逆序对,用树状数组保存l1的信息,对于每个l2,查询前缀和,然后删除自身

注意细节

#include<bits/stdc++.h>
#define endl '
'
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
#define mp make_pair
#define pii pair<ll,int>
#define all(x) x.begin(),x.end()
#define IO ios::sync_with_stdio(false)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(a,b) cout<<#a<<'['<<b<<"]="<<a[b]<<endl
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define forn(i,x) for(int i=head[x];i;i=e[i].next)
using namespace std;
const int maxn=1e5+10,maxm=2e5+10;
const int INF=0x3f3f3f3f;
const int mod=1e9+7;
const double PI=acos(-1.0);
int casn,n,m;
ll k;
struct bit{
    ll node[maxn];
    inline int lb(int x) {return x&(-x);}
    inline void update(int pos,ll val=1){
        pos++;if(pos<=0) return ;
        for(int i=pos;i&&i<=n;i+=lb(i))node[i]+=val;
    }
    inline ll ask(int pos){
        ll sum=0;pos++;if(pos<=0) return 0;
        for(int i=pos;i>0;i-=lb(i))sum+=node[i];return sum;
    }
    inline ll query(int l,int r){return ask(r)-ask(l-1);}
}tree;
class graph{public:
  struct node{int to,next;ll cost;}e[maxm];
  int head[maxn],nume,n,sz[maxn],maxt,stree[maxn];
  void add(int a,int b,ll c=0){e[++nume]={b,head[a],c};head[a]=nume;}
  int vis[maxn],num[maxn],all,mid;
  void getmid(int now=1,int pre=0){
    sz[now]=1;
    for(int i=head[now];i;i=e[i].next){
      if(e[i].to==pre||vis[e[i].to]) continue;
      getmid(e[i].to,now);
      sz[now]+=sz[e[i].to];
    }
    int tmp=max(sz[now]-1,all-sz[now]);
    if(maxt>tmp) maxt=tmp,mid=now;
  }//base
  pii dis[maxn];
  int dfn,mxd;
  ll ans;
  void init(int n){
    this->n=n,nume=1,mid=0;
    rep(i,1,n) vis[i]=head[i]=0;
  }
  ll getdis(int now,int pre,int d,ll s){
    dis[++dfn]={s,d};mxd=max(mxd,d);
    for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(to==pre||vis[to]) continue;
      getdis(to,now,d+1,s+e[i].cost);
    }
  }
  ll getans(int now,int d=0,ll s=0){
    dfn=0;mxd=0;
    getdis(now,0,d,s);
    sort(dis+1,dis+1+dfn);
    ll ans=0;
    int l=1,r=dfn;
    while(l<r&&dis[l].fi+dis[r].fi>k)r--;
    rep(i,l+1,r) tree.update(dis[i].se);
    while(l<r){
        if(dis[l].fi+dis[r].fi<=k) {
            ans+=tree.ask(min(m-dis[l].se,mxd));
            tree.update(dis[++l].se,-1);
        }else tree.update(dis[r--].se,-1);
    }
    return ans;
  }
  void divide(int now){
    vis[now]=1;ans+=getans(now);
    for(int i=head[now];i;i=e[i].next){
      int to=e[i].to;
      if(vis[to]) continue;
      ans-=getans(to,1,e[i].cost);
      all=sz[to],maxt=n+1;
      getmid(to,now);divide(mid);
    }
  }
  void solve(){
    ans=0;maxt=all=n;
    getmid();divide(mid);
  }
}g;

int main() {IO;
  cin>>n>>m>>k;
  g.init(n);
  rep(i,2,n){
    int a;ll c;cin>>a>>c;
    g.add(i,a,c);g.add(a,i,c);
  }
  g.solve();
  cout<<g.ans<<endl;
}