题目链接：BZOJ 洛谷
\(O(n^2)\)DP很好写，对于当前的i从之前满足条件的j中选一个最大值,\(dp[i]=d[j]+1\)

 for(int j=1; j<i; ++j)if(a[j]<=minv[i]&&maxv[j]<=a[i])//序列只会变换一次 dp[i]=max{dp[j]+1};

转移要满足两个条件：\(a[j]<=minv[i]\ \&\&\ maxv[j]<=a[i]\)
一个二维偏序问题，CDQ、树套树都可以。
拿\(minv[x]\)和\(a[y]\)作为两个坐标轴，\(dp[j]\)表示其上一点\((A[j],maxv[j])\).
这样就成了一个二维平面，可以向其中插入一些点dp[i]，询问一个矩形区域(也是一段前缀)中某点最大值
线段树套线段树 树状数组套线段树都可做 复杂度\(O(nlog^2n)\)
后者时间还可以
树状数组套线段树：

//企图二维树状数组 但动态开点的话 中间一段没有的区间会中断y方向的Query..
#include<cstdio>
#include<cctype>
#include<algorithm>
#define gc() getchar()
#define now node[rt]
#define lson l,m,node[rt].ls
#define rson m+1,r,node[rt].rs
#define lb(x) ((x)&-(x))
const int N=1e5+5;int n,m,A[N],minv[N],maxv[N],MaxV,MaxA;
namespace Tree_2D
{struct Seg_Tree{struct Node{int maxv,ls,rs;}node[N<<6];//还要再小点。。不然BZOJ上依旧MLE inline int new_Node(){static int cnt=0;return ++cnt;}void Insert(int l,int r,int &rt,int p,int v){if(!rt) rt = new_Node();now.maxv = std::max(now.maxv, v);if(l<r){int m=l+r>>1;if(p<=m) Insert(lson,p,v);else Insert(rson,p,v);}}int Query(int l,int r,int rt,int L,int R){if(!rt) return 0;if(L<=l && r<=R) return now.maxv;int m=l+r>>1;if(L<=m)if(m<R) return std::max(Query(lson,L,R),Query(rson,L,R));else return Query(lson,L,R);return Query(rson,L,R);}}t;struct Bit{int root[N];void Insert(int p,int y,int v){while(p<=MaxV)t.Insert(1,MaxA,root[p],y,v), p+=lb(p);}int Query(int p,int y){int res=0;while(p)res=std::max(res,t.Query(1,MaxA,root[p],1,y)), p-=lb(p);return res;}}t2D;
}
#undef now
inline int read()
{int now=0;register char c=gc();for(;!isdigit(c);c=gc());for(;isdigit(c);now=now*10+c-'0',c=gc());return now;
}int main()
{n=read(),m=read();for(int i=1; i<=n; ++i)maxv[i]=minv[i]=A[i]=read(), MaxA=std::max(MaxA,A[i]);for(int x,y,i=1; i<=m; ++i)x=read(), y=read(), maxv[x]=std::max(maxv[x],y), minv[x]=std::min(minv[x],y);for(int i=1; i<=n; ++i) MaxV=std::max(MaxV,maxv[i]);int ans=0;for(int v,i=1; i<=n; ++i){v = Tree_2D::t2D.Query(minv[i],A[i]) + 1;Tree_2D::t2D.Insert(A[i],maxv[i],v);ans=std::max(ans,v);}printf("%d",ans);return 0;
}

二维线段树：

/*
BZOJ上直接MLE...洛谷P4093 4508ms(比Bit套Segtree慢3倍+) 293.33MB
空间消耗比较大 写指针吧。。
*/
#include<cstdio>
#include<cctype>
#include<algorithm>
#define gc() getchar()
#define lson l,m,rt->ls
#define rson m+1,r,rt->rs
const int N=1e5+5;int n,m,A[N],maxv[N],minv[N],MaxA,MaxV;
namespace Seg_Tree2D
{struct Node{int maxv;Node *ls,*rs;Node(): maxv(0),ls(NULL),rs(NULL) { }}pool[N<<7];//(logN)^2=256(2^8) 开得小点吧要不空间会炸 struct Node2D{Node *root;Node2D *ls,*rs;Node2D(): root(NULL),ls(NULL),rs(NULL) { }}pool2D[N<<1],*root;inline Node *new_Node(){static int cnt=0;return &pool[cnt++];}inline Node2D *new_Node2D(){static int cnt=0;return &pool2D[cnt++];}Node2D *Build(int l,int r){Node2D *rt = new_Node2D();if(l<r){int m=l+r>>1;rt->ls = Build(l,m);rt->rs = Build(m+1,r);}return rt;}int Query(int l,int r,Node *rt,int L,int R){if(!rt) return 0;if(L<=l && r<=R) return rt->maxv;int m=l+r>>1;if(L<=m)if(m<R) return std::max(Query(lson,L,R),Query(rson,L,R));else return Query(lson,L,R);return Query(rson,L,R);}int Query2D(int l,int r,Node2D *rt,int L,int R,int y1,int y2){if(L<=l && r<=R) return Query(1,MaxA,rt->root,y1,y2);int m=l+r>>1;if(L<=m)if(m<R) return std::max(Query2D(lson,L,R,y1,y2),Query2D(rson,L,R,y1,y2));else return Query2D(lson,L,R,y1,y2);return Query2D(rson,L,R,y1,y2);}void Insert(int l,int r,Node *&rt,int p,int v){if(!rt) rt = new_Node();//!rt->maxv = std::max(rt->maxv, v);if(l<r){int m=l+r>>1;if(p<=m) Insert(lson,p,v);else Insert(rson,p,v);}}void Insert2D(int l,int r,Node2D *rt,int p,int y,int v){Insert(1, MaxA, rt->root, y, v);if(l<r){int m=l+r>>1;if(p<=m) Insert2D(lson,p,y,v);else Insert2D(rson,p,y,v);}}void Init(){root = Build(1,MaxV);}int Query_Max(int l,int r,int y1,int y2){return Query2D(1,MaxV,root,l,r,y1,y2);}void Insert_Node(int x,int y,int v){Insert2D(1,MaxV,root,x,y,v);}
}
inline int read()
{int now=0;register char c=gc();for(;!isdigit(c);c=gc());for(;isdigit(c);now=now*10+c-'0',c=gc());return now;
}int main()
{n=read(),m=read();for(int i=1; i<=n; ++i)maxv[i]=minv[i]=A[i]=read(), MaxA=std::max(MaxA,A[i]);for(int x,y,i=1; i<=m; ++i)x=read(), y=read(), maxv[x]=std::max(maxv[x],y), minv[x]=std::min(minv[x],y);for(int i=1; i<=n; ++i) MaxV=std::max(MaxV,maxv[i]);Seg_Tree2D::Init();int ans=0;for(int v,i=1; i<=n; ++i){v = Seg_Tree2D::Query_Max(1,minv[i],1,A[i]) + 1;Seg_Tree2D::Insert_Node(A[i],maxv[i],v);ans=std::max(ans,v);}printf("%d",ans);return 0;
}