bzoj 1264

用DP直接求LCS明显超时，这题和平时的区别在与给定了数，所以从这入手可以直接记录各数字的位置

然后依次填入dp[i]=max{dp[j]}+1(1<=j<=i)用BIT维护最大值O(nlogn)

//#include<bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<iostream>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll long long
#define succ(x) (1<<x)
#define lowbit(x) (x&(-x))
#define NM 20000+5
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
    return x*f;
}
int n,c[5*NM],_x,b[NM][6],ans;
void upd(int x,int t){
    for(;x<=5*n;x+=lowbit(x))c[x]=max(c[x],t);
}
int sum(int x){
    int s=0;
    for(;x;x-=lowbit(x))s=max(s,c[x]);
    return s;
}
int main(){
    freopen("data.in","r",stdin);
    n=read();
    inc(i,1,5*n){
        _x=read();b[_x][++b[_x][0]]=i;
    }
    inc(i,1,5*n){
        _x=read();
        dec(j,5,1){
            int t=b[_x][j];
            upd(t,sum(t-1)+1);
        }
    //    inc(i,1,5*n)printf("%d ",c[i]);printf("
");
    }
    printf("%d
",sum(5*n));
    return 0;
}