Partial Credit: Quadratic Time

When $N\le 1000$, we can afford to iterate over all pairs of boxes.

A box $i$ is blocked by another box $j$ ($j\neq i$) if $(x_{j1}, y_{j1})$ is to the lower-left of $(x_{i2},y_{i2})$. Consider a graph with directed edges $i\to j$ for all such pairs of boxes.

$M=2$: For each $i$, check if there is an edge to some $j>i$. If so, output $\texttt{0}$. Otherwise, output $\texttt{1}$.

$M=1$: Construct a topological ordering of the graph, and then output its reversal. The fact that a valid order exists is equivalent to the graph being acyclic.

Both $M=1$ and $M=2$ run in $O(N^2)$ time (and $O(N)$ memory, if you don't explicitly store the edges).

Full Credit: Segment Tree

Note that there are some solutions for $M=1$ only, but because they do not generalize to $M=2$, they are described after this section.

The graph described above has too many edges, so we want to optimize the key operations we perform on it. At its core, we would like a data structure supporting the following operations efficiently:

1. Update: Deactivate a box.
2. Query: Given an active box, output another active box it is blocked by, or determine that no such box exists.

We can support these operations in $O(\log N)$ time each using a segment tree (see: abstract description) storing the lower-left corners of each active box, supporting range minimum queries. Specifically, define an integer array $a$ with all entries initially set to $\infty$, and then for each active box $i$ set $a[x_{i1}]=y_{i1}$. An Update corresponds to updating $a[x_{i1}] = \infty$. For a Query, an active box $i$ is blocked if after $i$ is deactivated, $\min(a[0\dots x_{i2}-1])<y_{i2}$. (Note we can easily re-activate box $i$ by updating $a[x_{i1}]=y_{i1}$.)

$M=2$: For each box in order, make a Query and then an Update. Output $\texttt{1}$ if the Query returns non-blocked, and $\texttt{0}$ otherwise.

$M=1$: We can use the DFS-based recursive implementation of topological sort (example implementation here), except that instead of looping over all $u$ out of $v$, we can use a Query to directly find a $u$ that has not yet been pushed to the answer list. This also means we should Update a box when it gets pushed, to remove it from consideration. Since the graph is acyclic, we will never iterate over the same vertex more than once, meaning that the total number of vertices iterated over is at most $N$.

Both $M=1$ and $M=2$ run in $O(N\log N)$ time and $O(N)$ memory.

Benjamin Qi's code:

#include <bits/stdc++.h>
using namespace std;

template <class T> using V = vector<T>;

struct S {  // y-coordinate and the box associated with it
    int y;
    int label;
};

S op(S a, S b) {  // combiner function: return the lower y-coordinate
    if (a.y < b.y) return a;
    return b;
}

S e() { return {(int)1e9, -1}; }  // identity


struct Segtree {  // simplified version of AtCoder's segment tree implementation
    int n = 1;
    V<S> d;
    Segtree(int n_) {
        while (n < n_) n *= 2;
        d = V<S>(2 * n, e());
    }
    void set(int p, S x) {
        d.at(p += n) = x;
        while (p > 1) p /= 2, d.at(p) = op(d.at(2 * p), d.at(2 * p + 1));
    }
    S prod(int l, int r) {
        assert(0 <= l && l <= r && r <= n);
        S vl = e(), vr = e();
        for (l += n, r += n; l < r; l /= 2, r /= 2) {
            if (l & 1) vl = op(vl, d.at(l++));
            if (r & 1) vr = op(d.at(--r), vr);
        }
        return op(vl, vr);
    }
};

int M;

void solve() {
    int N;
    cin >> N;
    V<array<array<int, 2>, 2>> boxes(N);
    for (auto &a : boxes)  // read and zero-index
        for (auto &b : a)
            for (auto &c : b) {
                cin >> c;
                --c;
            }
    Segtree seg(2 * N);
    auto rem = [&](int x) {  // deactivate op
        seg.set(boxes.at(x).at(0).at(0), e());
    };
    auto add = [&](int x) {  // activate op
        seg.set(boxes.at(x).at(0).at(0), S{boxes.at(x).at(0).at(1), x});
    };
    auto edge_out = [&](int i) {  // query op: returns -1 if not blocked, or
                                  // some blocking box
        rem(i);
        S s = seg.prod(0, boxes.at(i).at(1).at(0));
        int ret = (s.y < boxes.at(i).at(1).at(1)) ? s.label : -1;
        add(i);
        return ret;
    };
    for (int i = 0; i < N; ++i) add(i);
    if (M == 1) {
        V<bool> active(N, true);
        V<int> ans;
        function<void(int)> dfs = [&](int x) -> void {
            assert(active.at(x));
            while (true) {
                int y = edge_out(x);
                if (y == -1) {
                    active.at(x) = false;
                    rem(x);
                    ans.push_back(x);
                    return;
                }
                dfs(y);
            }
        };
        for (int i = 0; i < N; ++i)
            if (active.at(i)) dfs(i);
        assert(ans.size() == N);
        for (int i = 0; i < N; ++i) {
            if (i) cout << " ";
            cout << ans[i] + 1;
        }
        cout << "\n";
    } else {
        string ans;
        for (int i = 0; i < N; ++i) {
            int x = edge_out(i);
            ans += (x == -1 ? '1' : '0');
            rem(i);
        }
        cout << ans << "\n";
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T;
    cin >> T >> M;
    while (T--) solve();
}

Alternative solutions for M=1

We would like some more lightweight version of the graph described above, by excluding unnecessary edges. The key insight is that while it is hard to construct a sparse graph where every topological sort forms a valid removal order, we can instead construct a sparse graph that stores only some information about how boxes block each other and use a specific toposort that accounts for the nature of the remaining blocking constraints.

One method is: From left to right, at every box's right side, add an edge to the nearest box directly below this right side (or an abstract "root" node if no boxes are below). This can be done with a sweep line. This forms a tree, and its pre-order traversal is a valid removal order.

Another method is: At every box's left side, add an edge to the nearest rectangles above and below this left side, if they exist. This can also be done with a sweep line. Toposort using Kahn's algorithm (keeping track of indegree), always removing the leftmost available box (by either $x$-coordinate).

To prove correctness, it suffices to show that if a box blocks another box, it will come earlier in the outputted order.

Unfortunately, these methods do not generalize to $M=2$, since they rely on specific ways to construct a valid ordering that are, in a sense, more restrictive than what is allowed. Notice that, in both methods above, only one $y$-coordinate of each box needs to be known!

Austin Geng's code for the latter method:

#include <bits/stdc++.h>
 
using namespace std;
 
template<typename T>
using PQ = priority_queue<T, vector<T>, greater<T>>;
 
int main() {
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
 
    int t; int m; cin>>t>>m;
    if (m != 1) {
        assert(false);
    }
    for (int ti = 0; ti < t; ++ti) {
        int n; cin>>n;
        vector<tuple<int,int,int>> events;
        vector<int> xs(n);
        for (int i = 0; i < n; ++i) {
            int x1,y1,x2,y2; cin>>x1>>y1>>x2>>y2;
            xs[i] = x1;
            events.emplace_back(x1,y1,i);
            events.emplace_back(x2,y1,-1);
        }
        sort(events.begin(),events.end());
        map<int,int> sweep;
        vector<vector<int>> out(n);
        vector<int> deps(n,0);
        for (auto [_,y,i] : events) {
            if (i == -1) {    
                sweep.erase(y);
                continue;
            }
            auto it = sweep.lower_bound(y);
            if (it != sweep.begin()) {
                auto j = prev(it)->second;
                out[j].push_back(i);
                ++deps[i];
            }
            if (it != sweep.end()) {
                auto j = it->second;
                out[i].push_back(j);
                ++deps[j];
            }
            sweep[y] = i;
        }
        PQ<pair<int,int>> next;
        for (int i = 0; i < n; ++i) {
            if (deps[i] != 0) continue;
            next.push({xs[i],i});
        }
        while (!next.empty()) {
            auto [_,a] = next.top(); next.pop();
            cout << a+1 << " ";
            for (auto b : out[a]) {
                --deps[b];
                if (deps[b] != 0) continue;
                next.push({xs[b],b});
            }
        }
        cout << "\n";
    }
}

Existence of a valid removal order

Although not part of the problem, one might wonder why a valid removal order always exists. Here is a simple informal proof (idea by Benjamin Chen).

It suffices to show that there always exists a box that can be removed, since this fact can be repeatedly applied to produce a removal order. Imagine that "gravity" drags boxes down to a "floor" at $y=0$. Observe that if a box was blocked by another box, it will remain blocked. The leftmost box on the ground is, and thus was, removable.