First consider vertex $n+1$, which can only visit once, meaning at most one $i\rightarrow n+1$ and at most one $n+1\rightarrow i$.
Then, there only three cases:
- $…\rightarrow i\rightarrow n+1\rightarrow i+1\rightarrow…$
- $…\rightarrow n\rightarrow n+1$
- $n+1\rightarrow1\rightarrow…$
// compile: make data
// run: ./data < data.in
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#ifdef LOCAL
#include <debug/codeforces.h>
#define debug(x...) _debug_print(#x, x);
#define Debug(x...) _debug_print_format(#x, x);
std::ifstream terminal("/dev/tty");
#define PP cerr<<"\033[1;30mpause...\e[0m",terminal.ignore();
#else
#define debug(x...)
#define Debug(x...)
#define PP
#endif
template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;}
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define sz(v) ((int)(v).size())
#define print(...) print_(__VA_ARGS__);
#define FIND(a, x) ((find(a.begin(),a.end(),(x))!=a.end())?1:0)
#define cmin(x,...) x=min({(x),__VA_ARGS__})
#define cmax(x,...) x=max({(x),__VA_ARGS__})
#define INTMAX (int)(9223372036854775807)
#define INF (int)(1152921504606846976)
#define NaN (int)(0x8b88e1d0595d51d1)
#define double long double
#define int long long
#define uint unsigned long long
#define MAXN 200010
void solve() {
int n; cin >> n;
vector<int> a(n);
rep(i, 0, n) cin >> a[i];
if (a[n-1] == 0) {
rep(i, 0, n+1) cout << i+1 << " ";
cout << endl;
return;
}
if (a[0] == 1) {
cout << n+1;
rep(i, 0, n) cout << " " << i+1;
cout << endl;
return;
}
rep(i, 0, n-1) {
if (a[i] == 0 && a[i+1] == 1) {
rep(j, 0, i+1) cout << j+1 << " ";
cout << n+1;
rep(j, i+1, n) cout << " " << j+1;
cout << endl;
return;
}
}
cout << -1 << endl;
}
int32_t main() {
ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
int _; cin >> _;
while (_--) solve();
return 0;
}
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