[点分治] LA 7148 LRIP

题意就是要求一棵树上的最长不下降序列，同时不下降序列的最小值与最大值不超过D。

点分 记录经过重心结尾是多少的不降/不升的长度 然后用set维护一个单调的东西来查询 也可以用线段树维护

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include<set>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef pair<int,int> abcd;

inline char nc()
{
	static char buf[100000],*p1=buf,*p2=buf;
	if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
	return *p1++;
}

inline void read(int &x)
{
	char c=nc(),b=1;
	for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
	for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}

const int oo=1<<30;
const int N=100005;

struct edge{
	int u,v,next;
}G[N<<1];
int head[N],inum;

inline void add(int u,int v,int p){
	G[p].u=u; G[p].v=v; G[p].next=head[u]; head[u]=p;
}

int n,D,ans;
int val[N];

int size[N],del[N];
int minimum,rt,sum;

#define V G[p].v
inline void Root(int u,int fa){
	size[u]=1; int maximum=0;
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa && !del[V])
			Root(V,u),size[u]+=size[V],maximum=max(maximum,size[V]);
	maximum=max(maximum,sum-size[u]);
	if (maximum<minimum) minimum=maximum,rt=u;
}

int len,c[N]; 

set<abcd> Set1,Set2;
typedef set<abcd>::iterator ITER;

inline void dfs1(int u,int fa){
	if (val[u]>=c[len]) c[++len]=val[u]; else return;
	abcd tem=abcd(val[u],len);
	ITER it=Set2.lower_bound(abcd(val[u]-D,0));
	if (it!=Set2.end()){
		ans=max(ans,len+it->second-1);
	}
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa && !del[V])
			dfs1(V,u);
	len--;
}

inline void update1(int u,int fa){
	if (val[u]>=c[len]) c[++len]=val[u]; else return;
	ITER it=(Set1.insert(abcd(val[u],len))).first,tem;
	int flag=0;
	if (it!=Set1.begin())
	{
		tem=it; tem--;
		if (tem->second>it->second) Set1.erase(it),flag=1;
	}
	while (!flag)
	{
		tem=it; tem++; if (tem==Set1.end()) break;
		if (tem->second<it->second) Set1.erase(tem);
		else break; 
	}
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa && !del[V])
			update1(V,u);
	len--;
}

inline void dfs2(int u,int fa){
	if (val[u]<=c[len]) c[++len]=val[u]; else return;
	abcd tem=abcd(val[u],len);
//	for (ITER it=Set1.begin();it!=Set1.end();it++)
//		printf("%d %d\n",it->first,it->second);
	ITER it=Set1.upper_bound(abcd(val[u]+D,1<<30));
	if (it!=Set1.begin()){
		it--; //printf("%d %d\n",it->first,it->second);
		ans=max(ans,len+it->second-1);
	}
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa && !del[V])
			dfs2(V,u);
	len--;
}

inline void update2(int u,int fa){
	if (val[u]<=c[len]) c[++len]=val[u]; else return;
	ITER it=(Set2.insert(abcd(val[u],len))).first,tem;
	int flag=0;
	tem=it; tem++;
	if (tem!=Set2.end() && tem->second>it->second)
		Set2.erase(it),flag=1; 
	while (!flag && it!=Set2.begin())
	{
		tem=it; tem--;
		if (tem->second<it->second) Set2.erase(tem);
		else break; 
	}
	for (int p=head[u];p;p=G[p].next)
		if (V!=fa && !del[V])
			update2(V,u);
	len--;
}

inline void Solve(int u)
{
	del[u]=1;
	Set1.clear(); Set2.clear(); 
	Set1.insert(abcd(val[u],1));
	Set2.insert(abcd(val[u],1));
	for (int p=head[u];p;p=G[p].next)
		if (!del[V])
		{
			c[len=1]=val[u]; dfs1(V,u);
			c[len=1]=val[u]; dfs2(V,u);
			c[len=1]=val[u]; update1(V,u);
			c[len=1]=val[u]; update2(V,u);
		}
	for (int p=head[u];p;p=G[p].next)
		if (!del[V])
		{
			minimum=n+1; sum=size[V]; Root(V,u);
			Solve(rt);
		}
}

int main()
{
	int T; int iu,iv;
	freopen("t.in","r",stdin);
	freopen("t.out","w",stdout);
	read(T);
	for (int _t=1;_t<=T;_t++)
	{
		read(n); read(D);
		for (int i=1;i<=n;i++) read(val[i]);
		for (int i=1;i<n;i++)
			read(iu),read(iv),add(iu,iv,++inum),add(iv,iu,++inum);
		ans=1;
		sum=n; minimum=n+1; Root(1,0);
		Solve(rt); 
		printf("Case #%d: %d\n",_t,ans);
		cl(head); inum=0;
		cl(del);
	}
	return 0;
}