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 to bi. 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).
题意：给你n个数，有q个询问，每一次替换c上的位置为d，问替换后的最长严格上升子序列的长度是多少。
思路：如果普通的替换再查找肯定超时了，所以我们要用线段树来处理。我们设f[i],g[i]分别为以i位置为尾点和起始点的最长严格上升子序列的长度，这个可以用线段树O(nlogn)的复杂度求出来，记录最长上升子序列的长度为maxlen,然后我们再设f1[i],f2[i]表示询问i替换后，询问替换的位置为c，以c位置为尾点和起始点的最长上升子序列长度。接下来我们要判断替换的节点是不是"关键点","关键点"的意思是，如果原来序列没有这个位置的点，那么原来序列的最长上升子序列的长度达不到maxlen。那么这个要怎么判断呢，我们可以开一个数组cnt[i],表示对于一个节点j，以j为尾节点的最长上升子序列的长度为i，且f[j]+g[j]-1==maxlen的这样符合条件的j点的个数总和。对于每一个位置，先判断f[i]+g[i]-1是不是等于maxlen，如果等于maxlen,那么我们就把cnt[f[i]]++。然后对于每一个询问，ans=max(是不是为关键点?maxlen:maxlen-1 ,f1[i]+g1[i]-1 )。
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<bitset>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef long double ldb;
#define inf 99999999
#define pi acos(-1.0)
#define maxn 400050
vector<pair<int,int> >ques[maxn]; //<idx,num>
int ans[maxn],a[maxn],pos[2*maxn];
int f1[maxn],g1[maxn],f[maxn],g[maxn];
int cnt[maxn];
int c[maxn],d[maxn];
struct node1{
struct node{
int l,r,maxlen;
}b[8*maxn];
void build(int l,int r,int th)
{
int mid;
b[th].l=l;b[th].r=r;
b[th].maxlen=0;
if(l==r)return;
mid=(l+r)/2;
build(l,mid,th*2);
build(mid+1,r,th*2+1);
}
int question(int l,int r,int th)
{
int mid;
if(b[th].l==l && b[th].r==r){
return b[th].maxlen;
}
mid=(b[th].l+b[th].r)/2;
if(r<=mid)return question(l,r,th*2);
else if(l>mid)return question(l,r,th*2+1);
return max(question(l,mid,th*2),question(mid+1,r,th*2+1) );
}
void update(int idx,int num,int th)
{
int mid;
if(b[th].l==idx && b[th].r==idx){
b[th].maxlen=num;return;
}
mid=(b[th].l+b[th].r)/2;
if(idx<=mid)update(idx,num,th*2);
else update(idx,num,th*2+1);
b[th].maxlen=max(b[th*2].maxlen,b[th*2+1].maxlen);
}
}L,R;
int main()
{
int n,m,i,j,tot;
while(scanf("%d%d",&n,&m)!=EOF)
{
int tot=0;
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
tot++;pos[tot]=a[i];
}
for(i=1;i<=m;i++){
scanf("%d%d",&c[i],&d[i]);
ques[c[i] ].push_back(make_pair(i,d[i] ) );
tot++;pos[tot]=d[i];
}
sort(pos+1,pos+1+tot);
tot=unique(pos+1,pos+1+tot)-pos-1;
L.build(1,tot,1);
R.build(1,tot,1);
int maxlen,num,t;
maxlen=0;
for(i=1;i<=n;i++){
for(j=0;j<ques[i].size();j++){
t=lower_bound(pos+1,pos+1+tot,ques[i][j].second)-pos;
if(t==1)num=1;
else num=L.question(1,t-1,1)+1;
f1[ques[i][j].first ]=num;
}
t=lower_bound(pos+1,pos+1+tot,a[i])-pos;
if(t==1)f[i]=1;
else f[i]=L.question(1,t-1,1)+1;
L.update(t,f[i],1);
maxlen=max(maxlen,f[i]);
}
for(i=n;i>=1;i--){
for(j=0;j<ques[i].size();j++){
t=lower_bound(pos+1,pos+1+tot,ques[i][j].second)-pos;
if(t==tot)num=1;
else num=R.question(t+1,tot,1)+1;
g1[ques[i][j].first ]=num;
}
t=lower_bound(pos+1,pos+1+tot,a[i])-pos;
if(t==tot)g[i]=1;
else g[i]=R.question(t+1,tot,1)+1;
R.update(t,g[i],1);
}
for(i=1;i<=n;i++){
cnt[i]=0;
}
for(i=1;i<=n;i++){
if(f[i]+g[i]-1==maxlen){
cnt[f[i] ]++;
}
}
int ans;
for(i=1;i<=m;i++){
if(f[c[i] ]+g[c[i] ]-1==maxlen && cnt[f[c[i] ] ]==1 ){
ans=maxlen-1;
}
else ans=maxlen;
ans=max(f1[i]+g1[i]-1,ans );
printf("%d
",ans);
}
}
return 0;
}
/*
15 14
76 9 32 82 40 91 46 5 12 69 44 97 30 13 29
4 73
13 84
14 51
5 99
7 47
14 32
4 12
11 20
9 65
15 95
10 26
5 25
2 62
11 81
*/