【bzoj4990】[Usaco2017 Feb]Why Did the Cow Cross the Road II

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 acc

eptable 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 o

n each side), please help FJ determine the maximum number of crosswalks he can draw over his road, s

uch 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 <= 10^5。

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

题解

经典问题

由于是全排列，每个点最多只能连出去9条边

把能连到的点从大到小加到数组里

最长上升子序列（严格上升）

代码

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while (ch<'0'||ch>'9'){if (ch=='-')f=-1;ch=getchar();}
while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
int n,a[100005],b[100005],c[900005],ans,q[900005],tot;
int main()
{
n=read();
for (int i=1;i<=n;i++)
{
int x=read();a[x]=i;
}
for (int i=1;i<=n;i++)
{
int x=read(),m=0;
for (int j=x-4;j<=x+4;j++)if (j>0&&j<=n)
b[++m]=a[j];
sort(b+1,b+m+1,greater<int>());
for (int j=1;j<=m;j++) c[++tot]=b[j];
}
for (int i=1;i<=tot;i++)
{
int l=0,r=ans;
while (l!=r)
{
int mid=(l+r+1)>>1;
if (q[mid]<c[i]) l=mid;else r=mid-1;
}
l++;
ans=max(ans,l);
q[l]=c[i];
}
cout<<ans;
return 0;
}