Codeforces Round #550 (Div. 3)C. Two Shuffled Sequences

Strictly increasing sequence is a sequence of integers [x1<x2<⋯<xk]. And strictly decreasing sequence is a sequence of integers [y1>y2>⋯>yl]. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

They were merged into one sequence a. After that sequence a got shuffled. For example, some of the possible resulting sequences a for an increasing sequence [1,3,4] and a decreasing sequence [10,4,2] are sequences [1,2,3,4,4,10] or [4,2,1,10,4,3].

This shuffled sequence a is given in the input.

Your task is to find any two suitable initial sequences. One of them should be strictly increasing and the other one — strictly decreasing. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print “NO”.

Input
The first line of the input contains one integer n (1≤n≤2⋅105) — the number of elements in a.

The second line of the input contains n integers a1,a2,…,an (0≤ai≤2⋅105), where ai is the i-th element of a.

Output
If there is a contradiction in the input and it is impossible to split the given sequence a to increasing and decreasing sequences, print “NO” in the first line.

Otherwise print “YES” in the first line and any two suitable sequences. Note that the empty sequence and the sequence consisting of one element can be considered as increasing or decreasing.

In the second line print ni — the number of elements in the strictly increasing sequence. ni can be zero, in this case the increasing sequence is empty.

In the third line print ni integers inc1,inc2,…,incni in the increasing order of its values (inc1<inc2<⋯<incni) — the strictly increasing sequence itself. You can keep this line empty if ni=0 (or just print the empty line).

In the fourth line print nd — the number of elements in the strictly decreasing sequence. nd can be zero, in this case the decreasing sequence is empty.

In the fifth line print nd integers dec1,dec2,…,decnd in the decreasing order of its values (dec1>dec2>⋯>decnd) — the strictly decreasing sequence itself. You can keep this line empty if nd=0 (or just print the empty line).

ni+nd should be equal to n and the union of printed sequences should be a permutation of the given sequence (in case of “YES” answer).

Examples
input
7
7 2 7 3 3 1 4
output
YES
2
3 7
5
7 4 3 2 1
input
5
4 3 1 5 3
output
YES
1
3
4
5 4 3 1
input
5
1 1 2 1 2
output
NO
input
5
0 1 2 3 4
output
YES
0

5
4 3 2 1 0
题意：给定一个序列，将它拆成两部分，一个试严格递增，一个是严格递减。不能发拆成功就是输出“No”。（一个数字的序列和空虚了符号递增和递减要求）
思路：正向跑一遍，逆向一遍
参考代码：

#include <bits/stdc++.h>
#define N 200005
#define ll long long int
#define inf 0x3f3f3f3f
using namespace std;
ll A[N],B[N],C[N];
bool vis[N];
int main()
{
    ll  n;
    cin>>n;
    for(ll i=1; i<=n; i++)
        cin>>A[i];
    sort(A+1,A+n+1);
    B[0]=-1;
    C[0]=inf;
    ll sum1=0;
    for(ll i=1; i<=n; i++)
    {
        if(B[sum1]<A[i]&&!vis[i])
        {
            B[++sum1]=A[i];
            vis[i]=1;
        }
    }
    ll sum2=0;
    for(ll i=n; i>=1; i--)
    {
        if(C[sum2]>A[i]&&!vis[i])
        {
            C[++sum2]=A[i];
            vis[i]=1;
        }
    }
    if(sum1+sum2==n)
    {
        cout<<"YES\n";
        cout<<sum2<<endl;
        for(ll i=sum2; i>=1; i--)
            if(i==sum2)
                cout<<C[i];
            else
                cout<<" "<<C[i];
        cout<<endl;
        cout<<sum1<<endl;
        for(ll i=sum1; i>=1; i--)
            if(i==sum1)
                cout<<B[i];
            else
                cout<<" "<<B[i];
        cout<<endl;
    }
    else
        cout<<"NO\n";
    return 0;
}