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)

 1 //#include<bits/stdc++.h>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<queue>
 6 #include<iostream>
 7 #define inc(i,l,r) for(int i=l;i<=r;i++)
 8 #define dec(i,l,r) for(int i=l;i>=r;i--)
 9 #define link(x) for(edge *j=h[x];j;j=j->next)
10 #define mem(a) memset(a,0,sizeof(a))
11 #define ll long long
12 #define succ(x) (1<<x)
13 #define lowbit(x) (x&(-x))
14 #define NM 20000+5
15 using namespace std;
16 int read(){
17     int x=0,f=1;char ch=getchar();
18     while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
19     while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
20     return x*f;
21 }
22 int n,c[5*NM],_x,b[NM][6],ans;
23 void upd(int x,int t){
24     for(;x<=5*n;x+=lowbit(x))c[x]=max(c[x],t);
25 }
26 int sum(int x){
27     int s=0;
28     for(;x;x-=lowbit(x))s=max(s,c[x]);
29     return s;
30 }
31 int main(){
32     freopen("data.in","r",stdin);
33     n=read();
34     inc(i,1,5*n){
35         _x=read();b[_x][++b[_x][0]]=i;
36     }
37     inc(i,1,5*n){
38         _x=read();
39         dec(j,5,1){
40             int t=b[_x][j];
41             upd(t,sum(t-1)+1);
42         }
43     //    inc(i,1,5*n)printf("%d ",c[i]);printf("\n");
44     }
45     printf("%d\n",sum(5*n));
46     return 0;
47 }

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