[BZOJ2253][2010 Beijing wc]纸箱堆叠（dp+bit+cdq分治）

题目描述

传送门

题解

最长上升子序列+三维偏序问题

裸的dp人人会对吧O(n2)

考虑对序列进行cdq分治，每一次用左边的区间来更新右边的区间

按照x分治，每一次按照y排序，然后z用树状数组查询

需要注意的是，这里全部都是严格大于，也就是说分治的时候划分到左右两边的点不能出现x相等的情况，这样的话就预处理一下然后在选中点的时候搞一搞就行了

并且更新的顺序必须是：递归左区间，处理左区间对右区间的影响，递归右区间

代码

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define N 300005

int A,Mod,n,LSH,ans;
int lsh
,L
,R
,C
,f
,ch
;
struct data{int x,y,z,id;}q
,a
,b
;

int cmp(data a,data b)
{
return a.x<b.x||(a.x==b.x&&a.y<b.y)||(a.x==b.x&&a.y==b.y&&a.z<b.z);
}
int cmpy(data a,data b)
{
return a.y<b.y;
}
void cover(int loc)
{
for (int i=loc;i<=LSH;i+=i&-i)
C[i]=0;
}
void add(int loc,int val)
{
for (int i=loc;i<=LSH;i+=i&-i)
C[i]=max(C[i],val);
}
int query(int loc)
{
int ans=0;
for (int i=loc;i>=1;i-=i&-i)
ans=max(ans,C[i]);
return ans;
}
void cdq(int l,int r)
{
if (q[l].x==q[r].x) return;
int mid=(l+r)>>1;
if (L[mid]!=l) mid=L[mid]-1;
else mid=R[mid];
cdq(l,mid);

int tot=0;
int pa=1,pb=1,acnt=0,bcnt=0;
for (int i=l;i<=mid;++i) a[++acnt]=q[i];
for (int i=mid+1;i<=r;++i) b[++bcnt]=q[i];
sort(a+1,a+acnt+1,cmpy);sort(b+1,b+bcnt+1,cmpy);
while (pb<=bcnt)
{
while (pa<=acnt&&a[pa].y<b[pb].y)
{
add(a[pa].z,f[a[pa].id]);
ch[++tot]=a[pa].z;
++pa;
}
f[b[pb].id]=max(f[b[pb].id],query(b[pb].z-1)+1);
++pb;
}
for (int i=1;i<=tot;++i) cover(ch[i]);

cdq(mid+1,r);
}
int main()
{
scanf("%d%d%d",&A,&Mod,&n);q[0].z=1;
for (int i=1;i<=n;++i)
{
q[i].x=(long long)q[i-1].z*A%Mod,q[i].y=(long long)q[i].x*A%Mod;
q[i].z=(long long)q[i].y*A%Mod;
lsh[++LSH]=q[i].x,lsh[++LSH]=q[i].y,lsh[++LSH]=q[i].z;
}
for (int i=1;i<=n;++i)
{
if (q[i].x>q[i].y) swap(q[i].x,q[i].y);
if (q[i].y>q[i].z) swap(q[i].y,q[i].z);
if (q[i].x>q[i].y) swap(q[i].x,q[i].y);
}
sort(lsh+1,lsh+LSH+1);LSH=unique(lsh+1,lsh+LSH+1)-lsh-1;
for (int i=1;i<=n;++i)
q[i].x=lower_bound(lsh+1,lsh+LSH+1,q[i].x)-lsh,
q[i].y=lower_bound(lsh+1,lsh+LSH+1,q[i].y)-lsh,
q[i].z=lower_bound(lsh+1,lsh+LSH+1,q[i].z)-lsh;
sort(q+1,q+n+1,cmp);
for (int i=1;i<=n;++i) q[i].id=i;

for (int i=1;i<=n;++i)
if (i==1||q[i].x!=q[i-1].x) L[i]=i;
else L[i]=L[i-1];
for (int i=n;i>=1;--i)
if (i==n||q[i].x!=q[i+1].x) R[i]=i;
else R[i]=R[i+1];
f[q[1].id]=1;
cdq(1,n);
for (int i=1;i<=n;++i) ans=max(ans,f[i]);
printf("%d\n",ans);
}