Codeforces 587C&588E Duff in the Army 树剖合并数组

题意

洛谷翻译

题解

最多询问编号最小的$10$个人.
考虑树剖,线段树节点储存$10$个最小的编号,向上推的时候暴力合并.
不过不能太暴力,也不能用$vector$合并,会T飞,要手写数组合并.
如果用主席树的话可以少一个log.

#include<bits/stdc++.h> //Ithea Myse Valgulious
namespace chtholly{
    
    
typedef long long ll;
#define re0 register int
#define rel register ll
#define rec register char
#define gc getchar
//#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<23,stdin),p1==p2)?-1:*p1++)
#define pc putchar
#define p32 pc(' ')
#define pl puts("")
/*By Citrus*/
char buf[1<<23],*p1=buf,*p2=buf;
inline int read(){
    
    
  int x=0,f=1;char c=gc();
  for (;!isdigit(c);c=gc()) f^=c=='-';
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return f?x:-x;
  }
template <typename mitsuha>
inline bool read(mitsuha &x){
    
    
  x=0;int f=1;char c=gc();
  for (;!isdigit(c)&&~c;c=gc()) f^=c=='-';
  if (!~c) return 0;
  for (;isdigit(c);c=gc()) x=(x<<3)+(x<<1)+(c^'0');
  return x=f?x:-x,1;
  }
template <typename mitsuha>
inline int write(mitsuha x){
    
    
  if (!x) return 0&pc(48);
  if (x<0) pc('-'),x=-x;
  int bit[20],i,p=0;
  for (;x;x/=10) bit[++p]=x%10;
  for (i=p;i;--i) pc(bit[i]+48);
  return 0;
  }
inline char fuhao(){
    
    
  char c=gc();
  for (;isspace(c);c=gc());
  return c;
  }
}using namespace chtholly;
using namespace std;
const int yuzu=1e5;
typedef int fuko[yuzu|10];
typedef vector<int> yudachi[yuzu|10];
yudachi lj,city;
struct vct{
    
      //手写合并用数组
int ans[25];
vct(){
    
    memset(ans,0,sizeof ans);}
void pb(int x){
    
    ans[++*ans]=x;}
int size(){
    
    return *ans;}
vct operator +(vct b) {
    
    
  vct llx;
  int i=1,j=1; 
  for (;i<=*ans&&j<=b.size()&&llx.size()<10;)
    llx.pb(ans[i]<b.ans[j]?ans[i++]:b.ans[j++]);
  for (;i<=*ans&&llx.size()<10;)
    llx.pb(ans[i++]);
  for (;j<=b.size()&&llx.size()<10;)
    llx.pb(b.ans[j++]);
  return llx;
  // for (int i=1;i<=*ans;++i) llx.pb(ans[i]);
  // for (int i=1;i<=b.size();++i) llx.pb(b.ans[i]);
  // inplace_merge(llx.ans+1,llx.ans+*ans+1,llx.ans+llx.size()+1);
  // if (llx.size()>10) *llx.ans=10; 
  // for (int i=11;i<=llx.size();++i) llx.ans[i]=0;
  // return llx;
  }
};
int n,m,q,cnt;
namespace tree_chain_splitting{
    
    
fuko sz,son,fa,dep,dfn,ord,top;
void dfs1(int u,int f){
    
    
  sz[u]=1,dep[u]=dep[fa[u]=f]+1;
  for (int v:lj[u]) if (v^f) {
    
    
    dfs1(v,u),sz[u]+=sz[v];
    if (sz[v]>sz[son[u]]) son[u]=v;
    }
  }
void dfs2(int u,int _top) {
    
    
  top[u]=_top,ord[dfn[u]=++cnt]=u;
  if (son[u]) dfs2(son[u],_top);
  for (int v:lj[u]) if (v^fa[u]&&v^son[u]) dfs2(v,v);
  }

struct segtree{
    
    
#define le rt<<1
#define ri le|1
#define ls le,l,mid
#define rs ri,mid+1,r
vct val[yuzu<<2|13];  
void build(int rt=1,int l=1,int r=n) {
    
    
  if (l==r) {
    
    
    for (int i:city[ord[l]])
      val[rt].pb(i);
    return;
    }
  int mid=l+r>>1;
  build(ls),build(rs);
  val[rt]=val[le]+val[ri];
  }
vct query(int ql,int qr,int rt=1,int l=1,int r=n) {
    
    
  if (ql>r||qr<l) return vct(); 
  if (ql<=l&&qr>=r) return val[rt];
  int mid=l+r>>1;
  return query(ql,qr,ls)+query(ql,qr,rs);
  }
}my_;

vct query(int u,int v){
    
    
  vct zxy;
  for (;top[u]^top[v];u=fa[top[u]]) {
    
    
    if (dep[top[u]]<dep[top[v]]) swap(u,v);
    zxy=zxy+my_.query(dfn[top[u]],dfn[u]);
    }
  if (dep[u]>dep[v]) swap(u,v);
  return zxy+my_.query(dfn[u],dfn[v]);
  }

int mian() {
    
    
  int i,u,v,a;
  read(n),read(m),read(q);
  for (i=1;i<n;++i) {
    
    
    u=read(),v=read();
    lj[u].push_back(v);
    lj[v].push_back(u);
    }
  for (i=1;i<=m;++i) 
    if (city[a=read()].size()<10) city[a].push_back(i);
  dfs1(1,0),dfs2(1,1);
  my_.build();
  for (;q--;) {
    
    
    u=read(),v=read(),a=read();
    vct lxy=query(u,v);
    write(min(lxy.size(),a)),p32;
    for (i=1;i<=min(lxy.size(),a);++i) write(lxy.ans[i]),p32;pl;
    }
  }
}

int main(){
    
    
tree_chain_splitting::mian();
}

谢谢大家.