[Usaco2017 Feb]Why Did the Cow Cross the Road II (Platinum)

Description

Farmer John is continuing to ponder the issue of cows crossing the road through his farm, introduced in the preceding problem. He realizes that interaction between some pairs of breeds is actually acceptable if the breeds are friendly, a property that turns out to be easily characterized in terms of breed ID: breeds aa and bb are friendly if |a-b|≤4, and unfriendly otherwise. It is ok for cows to wander into fields designated for other breeds, as long as they are friendly.Given the ordering of N fields on both sides of the road through FJ's farm (again, with exactly one field for each breed on each side), please help FJ determine the maximum number of crosswalks he can draw over his road, such that no two intersect, and such that each crosswalk joins a pair of fields containing two breeds that are friendly. Each field can be accessible via at most one crosswalk (so crosswalks don't meet at their endpoints).

上下有两个长度为n、位置对应的序列A、B，其中数的范围均为1~n。若abs(A[i]-B[j]) <= 4，则A[i]与B[j]间可以连一条边。现要求在边与边不相交的情况下的最大的连边数量。n <= 10^6

Input

The first line of input contains N (1≤N≤100,0000).

The next N lines describe the order, by breed ID, of fields on one side of the road;

each breed ID is an integer in the range 1…N The last N lines describe the order, by breed ID, of the fields on the other side of the road.

Each breed ID appears exactly once in each ordering.

注意:两个序列都是全排列

Output

Please output the maximum number of disjoint "friendly crosswalks" Farmer John can draw across the road.

Sample Input

6

1

2

3

4

5

6

6

5

4

3

2

1

Sample Output

5

---

首先可以想到二维dp，设\(f[i][j]\)表示序列A的前\(i\)个数和序列B的前\(j\)个数的最大连边数，那么就有

\[f[i][j]=max\{f[i-1][j],f[i][j-1],(f[i-1][j-1]+1)*[|A_i-B_j|<=4]\}
\]

但是这样转移会TLE，于是我们需要给\(f[i][j]\)增加一个限制条件，即\(A_i\)与\(B_j\)之间有连线，这样的话转移即为\(f[i][j]=max(f[x][y]+1),(1<x<i,1<y<j)\)，这样转移是需要前缀最大值的，于是就可以使用树状数组进行优化，而且第二维可以省去

/*program from Wolfycz*/
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x7f7f7f7f
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
inline int frd(){
	int x=0,f=1;char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline int read(){
	int x=0,f=1;char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)    putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=1e5;
int v[N+10],pos[N+10],now[N+10],tree[N+10],n;
void insert(int x,int v){for (;x<=n;x+=lowbit(x))	tree[x]=max(tree[x],v);}
int Query(int x){
	if (!x)	return 0;
	int res=0;
	for (;x;x-=lowbit(x))	res=max(res,tree[x]);
	return res;
}
int main(){
	n=read();
	for (int i=1;i<=n;i++)	v[i]=read();
	for (int i=1;i<=n;i++)	pos[read()]=i;
	for (int i=1;i<=n;i++){
		for (int j=max(1,v[i]-4);j<=min(n,v[i]+4);j++)	now[j]=Query(pos[j]-1);
		for (int j=max(1,v[i]-4);j<=min(n,v[i]+4);j++)	insert(pos[j],now[j]+1);
	}
	printf("%d\n",Query(n));
	return 0;
}