又是权限题qwq
一句话题意：带修改树上路径第k大
sol
数据结构？还是再见吧。学一手合格的整体二分，只有思维强大，才能见题拆题。
如果你做过整体二分的动态区间第k大就会发现这是一样的题。
无非是区间变成了树上路径。
怎么办呢？树剖+dfn序不就好了吗。
复杂度算下来貌似是\(O(n\log^3{n})\)的。
\(O(\mbox{跑得过})\)
code
注意是第k大啊！第k大啊！
写成第k小就爆零了。
神坑样例。
#include<cstdio>#include<algorithm>#include<cstring>#include<queue>using namespace std;const int N = 8e4+5;int gi(){ ???int x=0,w=1;char ch=getchar(); ???while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar(); ???if (ch=='-') w=0,ch=getchar(); ???while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar(); ???return w?x:-x;}struct edge{int to,next;}a[N<<1];struct node{int opt,x,y,id,k;}q[N<<2],q1[N<<2],q2[N<<2];int n,m,head[N],cnt,qcnt,T[N],fa[N],dep[N],sz[N],son[N],top[N],dfn[N],tmp[N],c[N],ans[N];void dfs1(int u,int f){ ???fa[u]=f;dep[u]=dep[f]+1;sz[u]=1; ???for (int e=head[u];e;e=a[e].next) ???{ ???????int v=a[e].to;if (v==f) continue; ???????dfs1(v,u); ???????sz[u]+=sz[v]; ???????if (sz[v]>sz[son[u]]) son[u]=v; ???}}void dfs2(int u,int up){ ???top[u]=up;dfn[u]=++cnt; ???if (son[u]) dfs2(son[u],up); ???for (int e=head[u];e;e=a[e].next) ???????if (a[e].to!=fa[u]&&a[e].to!=son[u]) ???????????dfs2(a[e].to,a[e].to);}void modify(int k,int v){while (k<=n) c[k]+=v,k+=k&-k;}int query(int k){int s=0;while (k) s+=c[k],k-=k&-k;return s;}int lca(int x,int y){ ???while (top[x]^top[y]) ???{ ???????if (dep[top[x]]<dep[top[y]]) swap(x,y); ???????x=fa[top[x]]; ???} ???return dep[x]<dep[y]?x:y;}int calc(int x,int y){ ???int res=0; ???while (top[x]^top[y]) ???{ ???????if (dep[top[x]]<dep[top[y]]) swap(x,y); ???????res+=query(dfn[x])-query(dfn[top[x]]-1); ???????x=fa[top[x]]; ???} ???if (dep[x]<dep[y]) swap(x,y); ???res+=query(dfn[x])-query(dfn[y]-1); ???return res;}void solve(int L,int R,int l,int r){ ???if (L>R) return; ???if (l==r) ???{ ???????for (int i=L;i<=R;i++) ???????????if (q[i].opt==0) ans[q[i].id]=l; ???????return; ???} ???int mid=l+r>>1; ???for (int i=L;i<=R;i++) ???????if (q[i].opt==0) tmp[q[i].id]=calc(q[i].x,q[i].y); ???????else if (q[i].y<=mid) modify(q[i].x,q[i].opt); ???for (int i=L;i<=R;i++) ???????if (q[i].opt!=0&&q[i].y<=mid) modify(q[i].x,-q[i].opt); ???int t1=0,t2=0; ???for (int i=L;i<=R;i++) ???????if (q[i].opt==0) ???????????if (q[i].k<=tmp[q[i].id]) q1[++t1]=q[i]; ???????????else q[i].k-=tmp[q[i].id],q2[++t2]=q[i]; ???????else ???????????if (q[i].y<=mid) q1[++t1]=q[i]; ???????????else q2[++t2]=q[i]; ???for (int i=L,j=1;j<=t1;i++,j++) q[i]=q1[j]; ???for (int i=L+t1,j=1;j<=t2;i++,j++) q[i]=q2[j]; ???solve(L,L+t1-1,l,mid);solve(L+t1,R,mid+1,r);}int main(){ ???n=gi();m=gi(); ???for (int i=1;i<=n;i++) T[i]=gi(); ???for (int i=1,u,v;i<n;i++) ???{ ???????u=gi(),v=gi(); ???????a[++cnt]=(edge){v,head[u]};head[u]=cnt; ???????a[++cnt]=(edge){u,head[v]};head[v]=cnt; ???} ???dfs1(1,0);cnt=0;dfs2(1,1);cnt=0; ???for (int i=1;i<=n;i++) ???????q[++cnt]=(node){1,dfn[i],T[i]}; ???while (m--) ???{ ???????int k=gi(),x=gi(),y=gi(); ???????if (k) ???????{ ???????????int len=dep[x]+dep[y]-2*dep[lca(x,y)]+1; ???????????if (k>len) k=1e8;else k=len-k+1; ???????????q[++cnt]=(node){0,x,y,++qcnt,k}; ???????} ???????else ???????{ ???????????q[++cnt]=(node){-1,dfn[x],T[x]}; ???????????T[x]=y; ???????????q[++cnt]=(node){1,dfn[x],T[x]}; ???????} ???} ???solve(1,cnt,0,1e8); ???for (int i=1;i<=qcnt;i++) ???????if (ans[i]==1e8) puts("invalid request!"); ???????else printf("%d\n",ans[i]); ???return 0;}