Codeforces Round #550 (Div. 3)C. Two Shuffled Sequences
Two integer sequences existed initially — one of them was strictly increasing, and the other one — strictly decreasing.
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
inputCopy
7
7 2 7 3 3 1 4
outputCopy
YES
2
3 7
5
7 4 3 2 1
inputCopy
5
4 3 1 5 3
outputCopy
YES
1
3
4
5 4 3 1
inputCopy
5
1 1 2 1 2
outputCopy
NO
inputCopy
5
0 1 2 3 4
outputCopy
YES
0
5
4 3 2 1 0
思路:稍微想一下就知道如果重复的数有3个或以上，那么必然不能分成两个序列。那么我们就用一个vector数组表示每一个数出现的个数。递增顺序就是从开始到结束每一个出现两次的数，剩下的就满足递减序列
#include <bits/stdc++.h>using namespace std;int main() {int n;cin >> n;vector<int> cnt(200 * 1000 + 1);for (int i = 0; i < n; ++i) {int x;cin >> x;++cnt[x];if (cnt[x] > 2) {cout << "NO" << endl;return 0;}}cout << "YES" << endl;cout<<count(cnt.begin(), cnt.end(), 2) << endl;for (int i = 0; i <= 200 * 1000; ++i) {if (cnt[i] == 2) {cout << i << " ";--cnt[i];}}cout << endl;cout<<count(cnt.begin(), cnt.end(), 1) << endl;for (int i = 200 * 1000; i >= 0; --i) {if (cnt[i] == 1) {cout << i << " ";--cnt[i];}}cout << endl;return 0;
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