2019.01.13 bzoj1146: [CTSC2008]网络管理Network（整体二分+树剖）

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题意简述：给一棵树，支持单点修改，询问路径上两点间第kkk大值。

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思路：

读懂题之后立马可以想到序列上带修区间kkk大数的整体二分做法，就是用一个bitbitbit来支持查值。

那么这个题把树状数组放到树上用树链剖分维护一下即可。

代码：

#include<bits/stdc++.h>
#define ri register int
using namespace std;
inline int read(){
	int ans=0;
	char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans;
}
const int N=8e4+5;
int n,m,a[N],siz[N],hson[N],fa[N],dep[N],top[N],num[N],pred[N],bit[N],ans[N],tot=0,sig=0,Q=0;
vector<int>e[N];
struct Node{int k,a,b,id;}q[N*3],qtmp1[N*3],qtmp2[N*3];
inline int lowbit(int x){return x&-x;}
inline void update(int x,int v){for(ri i=x;i<=n;i+=lowbit(i))bit[i]+=v;}
inline int query(int x){int ret=0;for(ri i=x;i;i-=lowbit(i))ret+=bit[i];return ret;}
void dfs1(int p){
	siz[p]=1;
	for(ri i=0,v;i<e[p].size();++i){
		if((v=e[p][i])==fa[p])continue;
		fa[v]=p,dep[v]=dep[p]+1,dfs1(v),siz[p]+=siz[v];
		if(siz[v]>siz[hson[p]])hson[p]=v;
	}
}
void dfs2(int p,int tp){
	top[p]=tp,pred[num[p]=++tot]=p;
	if(!hson[p])return;
	dfs2(hson[p],tp);
	for(ri i=0,v;i<e[p].size();++i)if((v=e[p][i])!=fa[p]&&v!=hson[p])dfs2(v,v);
}
inline 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;
}
inline int dist(int a,int b){return dep[a]+dep[b]-2*dep[lca(a,b)]+1;}
inline int ask(int x,int y){
	int ret=0;
	while(top[x]^top[y]){
		if(dep[top[x]]<dep[top[y]])swap(x,y);
		ret+=query(num[x])-query(num[top[x]]-1),x=fa[top[x]];
	}
	if(dep[x]<dep[y])swap(x,y);
	return ret+query(num[x])-query(num[y]-1);
}
inline void solve(int ql,int qr,int vl,int vr){
	if(ql>qr)return;
	if(vl==vr){for(ri i=ql;i<=qr;++i)if(q[i].k&&~ans[q[i].id])ans[q[i].id]=vl;return;}
	int mid=vl+vr>>1,hd1=0,hd2=0;
	for(ri i=ql;i<=qr;++i){
		if(!q[i].k){
			if(q[i].b<=mid)qtmp1[++hd1]=q[i];
			else qtmp2[++hd2]=q[i],update(num[q[i].a],q[i].id);
		}
		else{
			int ret=ask(q[i].a,q[i].b);
			if(ret>=q[i].k)qtmp2[++hd2]=q[i];
			else q[i].k-=ret,qtmp1[++hd1]=q[i];
		}
	}
	for(ri i=ql;i<=qr;++i)if(!q[i].k&&q[i].b>mid)update(num[q[i].a],-q[i].id);
	for(ri i=1;i<=hd1;++i)q[ql+i-1]=qtmp1[i];
	for(ri i=1;i<=hd2;++i)q[ql+hd1+i-1]=qtmp2[i];
	solve(ql,ql+hd1-1,vl,mid),solve(ql+hd1,qr,mid+1,vr);
}
int main(){
	n=read(),m=read();
	for(ri i=1;i<=n;++i)q[++sig]=(Node){0,i,a[i]=read(),1};
	for(ri i=1,u,v;i<n;++i)u=read(),v=read(),e[u].push_back(v),e[v].push_back(u);
	dfs1(1),dfs2(1,1);
	for(ri i=1,k,x,y;i<=m;++i){
		k=read(),x=read(),y=read();
		if(!k){
			q[++sig]=(Node){k,x,a[x],-1};
			q[++sig]=(Node){k,x,a[x]=y,1};
		}
		else{
			++Q;
			int dis=dist(x,y);
			if(dis<k)ans[Q]=-1;
			q[++sig]=(Node){k,x,y,Q};
		}
	}
	solve(1,sig,1,100000000);
	for(ri i=1;i<=Q;++i){
		if(~ans[i])cout<<ans[i]<<'\n';
		else puts("invalid request!");
	}
	return 0;
}