做法：ans=max(第i个位置空缺时的lis,第i个位置为x，以它结尾的lis+以它开头的lis-1)。
后者扫两遍即可求出。
前者只需要知道，lis中，第i个位置的数无可替代即可，若可替代就是lis的长度，否则是lis的长度-1。
如何判断？因为若无可替代，则当把这个数丢到辅助数组的某个位置后，它就不会再被后面的值替换。这样就做出来了。还是很有意思的一个题。
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
struct Q
{
int id,pos,val;
bool operator <(Q one)const
{
return pos<one.pos;
}
}q[400010];
int a[400010],b[400010],c[400010],d[400010],ans[400010];
bool isok[400010];
int n,m;
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)
scanf("%d",a+i);
for(int i=0;i<m;i++)
{
scanf("%d%d",&q[i].pos,&q[i].val);
q[i].pos--;
q[i].id=i;
}
sort(q,q+m);
int cnt=0,len=0;
for(int i=0;i<n;i++)
{
int index;
while(cnt<m&&q[cnt].pos==i)
{
index=lower_bound(b,b+len,q[cnt].val)-b;
ans[q[cnt].id]=index+1;
cnt++;
}
index=lower_bound(b,b+len,a[i])-b;
b[index]=a[i];
c[i]=index;
if(index==len)
len++;
}
int mx=len;
cnt=m-1,len=0;
memset(d,-1,sizeof(d));
for(int i=n-1;i>-1;i--)
{
int index;
while(cnt>-1&&q[cnt].pos==i)
{
index=lower_bound(b,b+len,-q[cnt].val)-b;
ans[q[cnt].id]+=index;
cnt--;
}
index=lower_bound(b,b+len,-a[i])-b;
b[index]=-a[i];
if(index+c[i]+1==mx)
{
if(d[c[i]]!=-1)
isok[d[c[i]]]=isok[i]=0;
else
{
d[c[i]]=i;
isok[i]=1;
}
}
if(index==len)
len++;
}
for(int i=0;i<m;i++)
ans[q[i].id]=max(ans[q[i].id],isok[q[i].pos]?mx-1:mx);
for(int i=0;i<m;i++)
printf("%d\n",ans[i]);
}Vasya has decided to build a zip-line on trees of a nearby forest. He wants the line to be as long as possible but he doesn't remember exactly the heights of all trees in the forest. He is sure that he remembers correct heights of all trees except, possibly, one of them.
It is known that the forest consists of n trees staying in a row numbered from left to right with integers from 1 to n. According to Vasya, the height of the i-th tree is equal to hi. The zip-line of length k should hang over k (1 ≤ k ≤ n) trees i1, i2, ..., ik (i1 < i2 < ... < ik) such that their heights form an increasing sequence, that is hi1 < hi2 < ... < hik.
Petya had been in this forest together with Vasya, and he now has q assumptions about the mistake in Vasya's sequence h. His i-th assumption consists of two integers ai and bi indicating that, according to Petya, the height of the tree numbered ai is actually equal tobi. Note that Petya's assumptions are independent from each other.
Your task is to find the maximum length of a zip-line that can be built over the trees under each of the q assumptions.
In this problem the length of a zip line is considered equal to the number of trees that form this zip-line.
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 400 000) — the number of the trees in the forest and the number of Petya's assumptions, respectively.
The following line contains n integers hi (1 ≤ hi ≤ 109) — the heights of trees according to Vasya.
Each of the following m lines contains two integers ai and bi (1 ≤ ai ≤ n, 1 ≤ bi ≤ 109).
For each of the Petya's assumptions output one integer, indicating the maximum length of a zip-line that can be built under this assumption.
4 4 1 2 3 4 1 1 1 4 4 3 4 5
4 3 3 4
4 2 1 3 2 6 3 5 2 4
4 3
Consider the first sample. The first assumption actually coincides with the height remembered by Vasya. In the second assumption the heights of the trees are (4, 2, 3, 4), in the third one they are (1, 2, 3, 3) and in the fourth one they are (1, 2, 3, 5).