Restore the Permutation by Sorted Segments CodeForces - 1343F greedy implementation

We guessed a permutation $p$

For each position $r$

For example, if the secret permutation is $p = [3, 1, 4, 6, 2, 5]$

$[2, 5, 6]$
$[4, 6]$
$[1, 3, 4]$
$[1, 3]$
$[1, 2, 4, 6]$

Your task is to find any suitable permutation (i.e. any permutation corresponding to the given input data). It is guaranteed that the input data corresponds to some permutation (i.e. such permutation exists).

You have to answer $t$

Input

The first line of the input contains one integer $t$

The first line of the test case contains one integer $n$

The next $n - 1$

The $i$

It is guaranteed that the required $p$

It is also guaranteed that the sum of $n$

Output

For each test case, print the answer: $n$

Example

Input

5
6
3 2 5 6
2 4 6
3 1 3 4
2 1 3
4 1 2 4 6
5
2 2 3
2 1 2
2 1 4
2 4 5
7
3 1 2 6
4 1 3 5 6
2 1 2
3 4 5 7
6 1 2 3 4 5 6
3 1 3 6
2
2 1 2
5
2 2 5
3 2 3 5
4 2 3 4 5
5 1 2 3 4 5

Output

3 1 4 6 2 5 
3 2 1 4 5 
2 1 6 3 5 4 7 
1 2 
2 5 3 4 1

tutorial：

Let's fix the first element and then try to restore permutation using this information. One interesting fact: if such permutation exists then it can be restored uniquely. Let's remove the first element from all segments containing it (we can use some logarithmic data structure for it). Then we just have a smaller problem but with one important condition: there is a segment consisting of one element (again, if such permutation exists). So, if the number of segments of length

If we succeed with restoring the permutation then we need to check if this permutation really satisfies the given input segments (see the first test case of the example to understand why this case appears). Let's just iterate over all

Time complexity:

//#include <bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cmath>
#include<cstring>
#include <algorithm>
#include <queue>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long ll;
const ll inf = 1<<30;
const int mod = 1000000007;
const int mx = 1e4+1e2; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(int i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(int i=(b),_=(a);i>=_;i--)
#define clr(a) memset(a, -1, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pai
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
int n, m, k,ans,t;

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);    
    cin >> t;
    
    while (t--) {
        cin >> n;
        vector<set<int>>se;
        re(i, 0, n - 1) {
            set <int>cur;
            int k;
            cin >> k;
            re(j, 0, k) {
                int x; cin >> x; cur.insert(x);
            }
            se.push_back(cur);
        }
        re(fst, 1, n + 1) {
            vector < int>ans;
            bool ok = 1;
            vector<set<int>>cur = se;
            for (auto& it : cur)if (it.count(fst))it.erase(fst);
            ans.push_back(fst);
            re(i, 1, n) {
                int cnt1 = 0, nxt = -1;
                for (auto& it : cur)if (it.size() == 1) {
                    ++cnt1; nxt = *it.begin();
                }
                if (cnt1 != 1) {
                    ok = 0; break;
                }
                for (auto& it : cur)if (it.count(nxt))it.erase(nxt);
                ans.push_back(nxt);
            }
            if (ok) {
                set<set<int>>all(se.begin(), se.end());
                re(i, 1, n) {
                    set<int>se;
                    se.insert(ans[i]);
                    bool find = 0;
                    for (int j = i-1; j >=0; --j)
                    {
                        se.insert(ans[j]);
                        if (all.count(se)) {
                            find = 1;
                            all.erase(se);
                            break;
                        }
                    }
                    if (!find)ok = 0;
                }
            }
            if (ok) {
                for (auto it : ans)cout << it << " ";
                cout << endl; break;
            }
        }
    }
    return 0;
}

Restore the Permutation by Sorted Segments CodeForces - 1343F greedy implementation

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