大家一起来做下这个典题。
我们很容易可以想到处理前后缀的最长上升子序列的长度,然后容易 \(O(n \log_2 n)\) 预处理。
做完了。
点击查看代码
/*
Tips:
你数组开小了吗?
你MLE了吗?
你觉得是贪心,是不是该想想dp?
一个小时没调出来,是不是该考虑换题?
打 cf 不要用 umap!!!
记住,rating 是身外之物。
该冲正解时冲正解!
Problem:
算法:
思路:
*/
#include<bits/stdc++.h>
using namespace std;
//#define map unordered_map
#define forl(i,a,b) for(register long long i=a;i<=b;i++)
#define forr(i,a,b) for(register long long i=a;i>=b;i--)
#define forll(i,a,b,c) for(register long long i=a;i<=b;i+=c)
#define forrr(i,a,b,c) for(register long long i=a;i>=b;i-=c)
#define lc(x) x<<1
#define rc(x) x<<1|1
#define mid ((l+r)>>1)
#define cin(x) scanf("%lld",&x)
#define cout(x) printf("%lld",x)
#define lowbit(x) (x&-x)
#define pb push_back
#define pf push_front
#define IOS ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
#define endl '\n'
#define QwQ return 0;
#define ll long long
#define ull unsigned long long
#define lcm(x,y) x/__gcd(x,y)*y
#define Sum(x,y) 1ll*(x+y)*(y-x+1)/2
#define aty cout<<"Yes\n";
#define atn cout<<"No\n";
#define cfy cout<<"YES\n";
#define cfn cout<<"NO\n";
#define xxy cout<<"yes\n";
#define xxn cout<<"no\n";
#define printcf(x) x?cout<<"YES\n":cout<<"NO\n";
#define printat(x) x?cout<<"Yes\n":cout<<"No\n";
#define printxx(x) x?cout<<"yes\n":cout<<"no\n";
ll t;
ll n,k,a[200010],b[200010],sum1[200010],sum2[200010],now,sum;
vector<ll>ans;
void solve()
{
cin>>n;
forl(i,1,n)
cin>>a[i];
k=0,b[++k]=a[1],sum1[1]=k;
forl(i,2,n)
{
if(b[k]<a[i])
b[++k]=a[i],sum1[i]=k;
else
now=lower_bound(b+1,b+k+1,a[i])-b,b[now]=a[i],sum1[i]=now;
}
sum=k,k=0,b[++k]=a[n]*-1,sum2[n]=k;
forr(i,n-1,1)
{
if(b[k]<a[i]*-1)
b[++k]=a[i]*-1,sum2[i]=k;
else
now=lower_bound(b+1,b+k+1,a[i]*-1)-b,b[now]=a[i]*-1,sum2[i]=now;
}
forl(i,1,n)
if(sum1[i]+sum2[i]==sum+1)
ans.pb(i);
cout<<ans.size()<<endl;
for(auto i:ans)
cout<<i<<' ';
ans.clear();
cout<<endl;
}
int main()
{
IOS;
t=1;
cin>>t;
while(t--)
solve();
/******************/
/*while(L<q[i].l) */
/* del(a[L++]);*/
/*while(L>q[i].l) */
/* add(a[--L]);*/
/*while(R<q[i].r) */
/* add(a[++R]);*/
/*while(R>q[i].r) */
/* del(a[R--]);*/
/******************/
QwQ;
}