题意:给定一颗树,每个叶子节点\(u\)都有权值\(val[u]\),求每个非叶子节点子树的最小叶子距离,若该子树只有一个叶子节点,输出INF

貌似本来是一道树分治(并不会)的题目,然而可以利用平衡树进行离线合并,边统计边更新
一开始没有想到这种方法,看了别人家的代码后觉得真是清晰明了
set交换后无需额外在dfs维护和叶节点更新后置为INF满足题意的单一叶子情况确实是不错的trick,值得学习

#include<bits/stdc++.h>
#define rep(i,j,k) for(register int i=j;i<=k;i++)
#define rrep(i,j,k) for(register int i=j;i>=k;i--)
#define erep(i,u) for(register int i=head[u];~i;i=nxt[i])
#define print(a) printf("%lld",(ll)(a))
#define println(a) printf("%lld\n",(ll)(a))
#define printbk(a) printf("%lld ",(ll)(a))
using namespace std;
const int MAXN = 5e4+11;
const int INF = 0x7fffffff;
typedef long long ll;
ll read(){ll x=0,f=1;register char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;
}
int to[MAXN<<1],nxt[MAXN<<1],head[MAXN],tot;
set<ll> s[MAXN];
ll n,m,val[MAXN];
void init(){memset(head,-1,sizeof head);tot=0;
}
void add(int u,int v){to[tot]=v;nxt[tot]=head[u];head[u]=tot++;swap(u,v);to[tot]=v;nxt[tot]=head[u];head[u]=tot++;
}
ll merge(set<ll> &a,set<ll> &b){if(a.size()<b.size()) swap(a,b);set<ll>::iterator it,lb,ub;ll mn=INF;for(it=b.begin();it!=b.end();it++){lb=ub=a.lower_bound(*it);if(lb!=a.begin()/*&&lb!=a.end()*/)lb--; //多了会WAif(lb!=a.end()) mn=min(mn,abs((*it)-(*lb)));if(ub!=a.end()) mn=min(mn,abs((*it)-(*ub)));a.insert(*it);//}b.clear();return mn;
}
#define isleaf(x) (bool((x)>=n-m+1&&(x)<=n))
void dfs(int u,int p){if(isleaf(u)){s[u].insert(val[u]);val[u]=INF;//trickreturn;}erep(i,u){int v=to[i];if(v==p)continue;dfs(v,u);val[u]=min(val[u],val[v]);val[u]=min(val[u],merge(s[u],s[v]));}
}
int main(){while(cin>>n>>m){init();rep(i,1,n) val[i]=INF;rep(i,1,n) s[i].clear();rep(i,2,n) add(read(),i);rep(i,n-m+1,n) val[i]=read();dfs(1,-1);rep(i,1,n-m){if(i==n-m) println(val[i]);else printbk(val[i]);}}return 0;
}