https://www.lydsy.com/JudgeOnline/problem.php?id=4553
一道比较难的题？可能是我CDQ很久没写的缘故了……
设l[i]~r[i]表示i数的变动范围，考虑f[i]=max(f[j])+1，其中j要满足：
j<i
a[j]<=l[i]
r[j]<=a[i]
很显然是CDQ三维（？）偏序的模型，但是有四个变量，所以归并排序貌似不可做……
改一下变量名变成：
tj<ti
xj<=yi
zj<=xi
我们一维排t，二维mid左右两边分别排x和y，然后按照顺序类似归并排序把两边的x和y合在一起，再用树状数组维护第三维就好啦！
当然注意每次我们处理完别忘了要回归原位！
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<cctype>
#include<vector>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=1e5+;
const int MAX=1e5;
inline int read(){
int X=,w=;char ch=;
while(!isdigit(ch)){w|=ch=='-';ch=getchar();}
while(isdigit(ch))X=(X<<)+(X<<)+(ch^),ch=getchar();
return w?-X:X;
}
struct node{
int x,y,z,t,ans;
}a[N];
int n,m,tr[N];
inline int lowbit(int x){return x&(-x);}
inline void add(int x,int y){
for(int i=x;i<=MAX;i+=lowbit(i))tr[i]=max(tr[i],y);
}
inline int query(int x){
int res=;
for(int i=x;i;i-=lowbit(i))res=max(res,tr[i]);
return res;
}
inline void mdy(int x){
for(int i=x;i<=MAX;i+=lowbit(i))tr[i]=;
}
inline bool cmpt(node a,node b){
return a.t<b.t;
}
inline bool cmpx(node a,node b){
return a.x==b.x?a.t<b.t:a.x<b.x;
}
inline bool cmpy(node a,node b){
return a.y==b.y?a.t<b.t:a.y<b.y;
}
void cdq(int l,int r){
if(l>=r)return;
int mid=(l+r)>>;
cdq(l,mid);
sort(a+l,a+mid+,cmpx);
sort(a+mid+,a+r+,cmpy);
for(int i=l,j=l,k=mid+;i<=r;i++){
if(j<=mid&&(k>r||a[j].x<=a[k].y))add(a[j].z,a[j].ans),j++;
else a[k].ans=max(a[k].ans,query(a[k].x)+),k++;
}
for(int i=l;i<=mid;i++)mdy(a[i].z);
sort(a+mid+,a+r+,cmpt);
cdq(mid+,r);
}
int main(){
n=read(),m=read();
for(int i=;i<=n;i++)a[i].x=a[i].y=a[i].z=read(),a[i].t=i;
for(int i=;i<=m;i++){
int x=read(),y=read();
a[x].y=min(a[x].y,y);
a[x].z=max(a[x].z,y);
}
for(int i=;i<=n;i++)a[i].ans=;
cdq(,n);
int ans=;
for(int i=;i<=n;i++)ans=max(ans,a[i].ans);
printf("%d\n",ans);
return ;
}
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+本文作者：luyouqi233。 +
+欢迎访问我的博客：http://www.cnblogs.com/luyouqi233/+
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