Let's suppose you have an array a, a stack s (initially empty) and an array b (also initially empty).
You may perform the following operations until both a and s are empty:
- Take the first element of a, push it into s and remove it from a (if a is not empty);
- Take the top element from s, append it to the end of array b and remove it from s (if s is not empty).
You can perform these operations in arbitrary order.
If there exists a way to perform the operations such that array b is sorted in non-descending order in the end, then array a is called stack-sortable.
For example, [3, 1, 2] is stack-sortable, because b will be sorted if we perform the following operations:
- Remove 3 from a and push it into s;
- Remove 1 from a and push it into s;
- Remove 1 from s and append it to the end of b;
- Remove 2 from a and push it into s;
- Remove 2 from s and append it to the end of b;
- Remove 3 from s and append it to the end of b.
After all these operations b = [1, 2, 3], so [3, 1, 2] is stack-sortable. [2, 3, 1] is not stack-sortable.
You are given k first elements of some permutation p of size n (recall that a permutation of size n is an array of size n where each integer from 1 to n occurs exactly once). You have to restore the remaining n - k elements of this permutation so it is stack-sortable. If there are multiple answers, choose the answer such that p is lexicographically maximal (an array q is lexicographically greater than an array p iff there exists some integer k such that for every i < k qi = pi, and qk > pk). You may not swap or change any of first k elements of the permutation.
Print the lexicographically maximal permutation p you can obtain.
If there exists no answer then output -1.
Input
The first line contains two integers n and k (2 ≤ n ≤ 200000, 1 ≤ k < n) — the size of a desired permutation, and the number of elements you are given, respectively.
The second line contains k integers p1, p2, ..., pk (1 ≤ pi ≤ n) — the first kelements of p. These integers are pairwise distinct.
Output
If it is possible to restore a stack-sortable permutation p of size n such that the first k elements of p are equal to elements given in the input, print lexicographically maximal such permutation.
Otherwise print -1.
Examples
5 3
3 2 1
3 2 1 5 4
5 3
2 3 1
-1
5 1
3
3 2 1 5 4
5 2
3 4
-1
题意：给你一个数N和一个数 k , 然后是长度为K的数组，
让你构造出一个N的全排列，使之前K项是给定的数组，并且满足这个全排列是stack-sortable
题目给了stack-sortable的定义。
思路：
可以通过折耳根stack-sortab的性质和stack的性质来完成本题。
首先我们要知道这题的一个关键点，当一个数插入到栈的条件是这个数x小于栈中所以的数。
那么我们首先对这K个数进行操作，对于每一个数p[i]，先判断能不能加到栈中（判断条件是栈为空或者比栈顶小），
不能加入到栈中的就一定是符合条件的，那么是直接输出-1.
加入到栈中之后，进行弹出操作，从1开始用一个变量来维护弹出到的最大数，对于栈顶就是能弹出的就先从栈中弹出。
扫完后对剩余的栈中元素进行操作，剩余的栈中元素只所以没有被弹出是因为肯定有some比它小的数在这K个中没出现。
那么我们就把（栈中元素之间）的数倒序分别输出
然后再把前K个没有的数进行倒序输出即可。
因为要求字典序最大，所以是倒序输出这些。
具体细节见accode
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), '\0', sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a,ll b){return b?gcd(b,a%b):a;} ll lcm(ll a,ll b){return a/gcd(a,b)*b;} ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;} inline void getInt(int* p); const int maxn=1000010; const int inf=0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ int n,k; int p[maxn]; int vis[maxn]; int main() { //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin); //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout); gbtb; cin>>n>>k; repd(i,1,k) { cin>>p[i]; vis[p[i]]++; } stack<int> st; int m=inf; int isok=1; int now=0; repd(i,1,k) { if(p[i]+1==now) { now++; }else { if(st.empty()) { st.push(p[i]); }else if(st.top()>p[i]) { st.push(p[i]); }else { isok=0; break; } while(st.size()&&st.top()==now+1) { st.pop(); now++; } } } if(!isok) { cout<<-1<<endl; return 0; } // while(!st.empty()&&st.size()!=1) // { // st.pop(); // } repd(i,1,k) { cout<<p[i]<<" "; } while(!st.empty()) { for (int i = st.top()-1;i>=1; --i) { if(!vis[i]) { cout<<i<<" "; vis[i]=1; }else { break; } /* code */ } now=max(now,st.top()); st.pop(); } for(int i=n;i>now;i--) { cout<<i<<" "; } cout<<endl; return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ' ' || ch == '\n'); if (ch == '-') { *p = -(getchar() - '0'); while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 - ch + '0'; } } else { *p = ch - '0'; while ((ch = getchar()) >= '0' && ch <= '9') { *p = *p * 10 + ch - '0'; } } }