<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

449. Dendrograms
Time limit per test: 0.5 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

Nod Ishimatsu is a student majoring computer science. He has wrote a program that performs hierarchical clustering, for the end-term project in a class of Introduction to Artificial Intellegence. Clustering is the process of grouping a set of objects into subsets called clusters, so that each cluster contains those similar in some sense. Hierarchical clustering is a way to do this indirectly; it produces trees (or in this problem collections of trees) called dendrograms. A dendrogram represents similarity among objects. The figure below shows an example dendrogram.


Leaves (at the bottom) denote objects. The horizontal lines split the objects into two or more subgroups and indicate objects in different subgroups are less similar than those in the same subgroup. Thus, basically, the upper the line is located, the less similar the objects or the sets of objects connected by it. We can have, for example, three clusters by using the top two horizontal lines. Nod's program also outputs dendrograms, but it has two problems. Firstly, output dendrograms are sometimes not fully connected, i.e. output can contain multiple trees. We don't deal with this problem here. Secondly, the produced dendrograms are fairly messy. Here is an example:


Maybe we want to change the order of the objects and have a clean diagram like the one shown above. Your task is to write a program for this purpose, i.e. reordering the objects for a clean diagram. Here, by clean diagrams, we mean those with no crosses of vertical and horizontal lines. There is usually more than one clean diagram; among them your program should pick the one in which the object of the smaller number comes to the left (in a lexicographical manner). Refer to the sections of Input and Output for the detailed specification.
Input
The input is given in the following format:
N M Q
splitInfo1
splitInfo2
...
splitInfoM
query1
query2
...
queryQ
The first line contains three integers N (1 ≤ N ≤ 100000), M (0 ≤ M < N), and Q (1 ≤ Q ≤ 1000, Q ≤ N), which represents the numbers of objects, horizontal lines, and queries respectively. Then M lines follow to describe horizontal lines. Each line is given in the following format: Y L V1 V2... VL Y (0 ≤ Y ≤ 109) denotes the vertical position of the horizontal line; the smaller value indicates the upper position (or the less similarity). L is the number of vertical lines (i.e. objects or subgroups of objects) connected by the line. Vi (for 1 ≤ i ≤ L) indicates each of the vertical lines. Each object is indicated by its identifier, a unique number from 1 to N. Each subgroup of objects is indicated by the identifier of an arbitrary object in the subgroup. Following the description of horizontal lines, there are Q lines each of which indicates a leaf index at which your program should report which object should be placed. The index is given by a number from 1 (leftmost) to N (rightmost).
Output
Print Q lines where the i-th line indicates the identifier of the object at the i-th queried position.
Example(s)
sample input
sample output
3 2 3
10 2 1 2
20 2 3 2
1
2
3
1
2
3

sample input
sample output
5 2 5
10 3 1 2 4
20 3 2 3 5
1
2
3
4
5
1
2
3
5
4

sample input
sample output
4 1 4
10 2 1 4
1
2
3
4
1
4
2
3

sample input
sample output
4 3 4
30 2 1 4
20 2 2 4
10 2 3 4
1
2
3
4
1
4
2
3

sample input
sample output
4 3 4
10 2 1 4
20 2 2 4
30 2 3 4
1
2
3
4
1
2
3
4

sample input
sample output
4 3 4
10 2 1 2
15 2 1 4
20 2 2 3
1
2
3
4
1
4
2
3

sample input
sample output
3 2 3
10 2 2 3
20 2 1 2
1
2
3
1
2
3

sample input
sample output
1 0 1
1
1

<|response|>
## 1. Abridged Problem Statement

We are given `N` objects and `M` horizontal lines of a possibly disconnected dendrogram.

Each horizontal line has:

```text
Y L V1 V2 ... VL
```

- `Y` is its vertical position. Smaller `Y` means higher in the dendrogram.
- The line connects `L` current subgroups.
- Each `Vi` is an object ID contained in one of those subgroups.

We need to reorder the `N` leaves so that the dendrogram can be drawn cleanly, meaning no crossings between vertical and horizontal lines.

There can be multiple clean orders. We must output the lexicographically smallest one, but only for `Q` queried positions.

Constraints:

```text
1 ≤ N ≤ 100000
0 ≤ M < N
1 ≤ Q ≤ 1000
```

---

## 2. Key Observations

### Observation 1: The dendrogram is a forest

Each object is a leaf.  
Each horizontal line is an internal node whose children are the subgroups connected by that line.

Because the dendrogram may be disconnected, the final structure is not necessarily one tree, but a forest.

---

### Observation 2: Process horizontal lines from bottom to top

A larger `Y` means the line is lower in the picture, meaning it connects more similar/smaller subgroups.

Therefore, when building the tree, we should process lines in decreasing `Y`.

When processing a line, all subgroups below it have already been built.

---

### Observation 3: Use DSU to know current subgroups

Initially, every object is its own subgroup.

As we process each horizontal line, it merges several current subgroups into a new subgroup.

A Disjoint Set Union structure can maintain these current subgroups.

For each DSU component, we store the current tree node representing that component.

---

### Observation 4: Clean drawings require contiguous subtrees

In a clean dendrogram drawing, every subtree must occupy a contiguous block of leaves.

Therefore, once the forest structure is fixed, the only freedom is:

1. The order of children of each internal node.
2. The order of roots of the forest.

---

### Observation 5: Lexicographically smallest order is obtained by sorting by subtree minimum

For every tree node `u`, define:

```text
subtree_min[u] = smallest object ID inside subtree u
```

Consider two sibling subtrees `A` and `B`.

The lexicographically smallest sequence of subtree `A` starts with `subtree_min[A]`.  
The lexicographically smallest sequence of subtree `B` starts with `subtree_min[B]`.

Since sibling subtrees contain disjoint leaves, these minimums are different.

So, to get the lexicographically smallest global order:

```text
Sort children by subtree_min.
Sort forest roots by subtree_min.
```

---

## 3. Full Solution Approach

### Step 1: Read and sort horizontal lines

Sort all horizontal lines by decreasing `Y`, because we want to process lower lines before upper lines.

---

### Step 2: Build the forest using DSU

We create tree nodes as follows:

- Leaf nodes are numbered `1` to `N`.
- Internal nodes are numbered `N + 1` to `N + M`.

Initially:

```text
DSU component i contains object i
node_of[i] = i
```

Here `node_of[root]` means:

```text
The current tree node representing DSU component root.
```

For each horizontal line connecting `V1, V2, ..., VL`:

1. Create a new internal node `cur`.
2. For each representative `Vi`:
   - Find its DSU component root `r`.
   - Add `node_of[r]` as a child of `cur`.
3. Union all these DSU components.
4. The merged component is now represented by `cur`.

---

### Step 3: Compute `subtree_min`

For leaves:

```text
subtree_min[i] = i
```

Internal nodes are created after their children, so every child has a smaller node ID than its parent.

Thus, we can compute:

```text
for u = N + 1 to N + M:
    subtree_min[u] = min(subtree_min[child] for child in children[u])
```

---

### Step 4: Sort children and roots

For each internal node:

```text
sort(children[u], by subtree_min)
```

After all lines are processed, every DSU root corresponds to one tree root in the final forest.

Collect those roots and sort them by `subtree_min`.

---

### Step 5: Traverse the forest

Perform an iterative DFS from the sorted roots.

Whenever we reach a leaf node, append it to the final leaf order.

Finally, answer each query `pos` with:

```text
order[pos - 1]
```

---

### Complexity

Let `S` be the total number of values `Vi` over all horizontal lines.

For a valid dendrogram, `S` is linear in `N`, because each merge reduces the number of connected components.

Complexities:

```text
Sorting lines:       O(M log M)
DSU operations:      O(S α(N))
Sorting children:    O(sum deg(u) log deg(u))
Traversal:           O(N + M)
```

Memory usage:

```text
O(N + M + S)
```

---

## 4. C++ Implementation with Detailed Comments

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

struct Line {
    int y;
    vector<int> members;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M, Q;
    cin >> N >> M >> Q;

    vector<Line> lines(M);

    for (int i = 0; i < M; i++) {
        int Y, L;
        cin >> Y >> L;

        lines[i].y = Y;
        lines[i].members.resize(L);

        for (int j = 0; j < L; j++) {
            cin >> lines[i].members[j];
        }
    }

    vector<int> queries(Q);
    for (int i = 0; i < Q; i++) {
        cin >> queries[i];
    }

    /*
        Larger Y means lower in the dendrogram.
        Lower lines should be processed first.
    */
    sort(lines.begin(), lines.end(), [](const Line& a, const Line& b) {
        return a.y > b.y;
    });

    /*
        Tree node IDs:
        1..N          : leaves / original objects
        N+1..N+M      : internal nodes / horizontal lines
    */
    int totalNodes = N + M;

    /*
        DSU parent array.
        Only original object indices 1..N are used in the DSU.
    */
    vector<int> parent(N + 1);

    /*
        nodeOf[root] stores the current tree node representing
        the DSU component whose root is root.
    */
    vector<int> nodeOf(N + 1);

    for (int i = 1; i <= N; i++) {
        parent[i] = i;
        nodeOf[i] = i;
    }

    /*
        children[u] stores the children of tree node u.
        Leaves have no children.
    */
    vector<vector<int>> children(totalNodes + 1);

    auto findRoot = [&](int x) {
        while (parent[x] != x) {
            parent[x] = parent[parent[x]];
            x = parent[x];
        }
        return x;
    };

    int nextNodeId = N;

    /*
        Build the dendrogram forest.
    */
    for (const Line& line : lines) {
        int cur = ++nextNodeId;

        /*
            Use the first member's component as the base component
            into which all other components will be merged.
        */
        int base = findRoot(line.members[0]);

        for (int v : line.members) {
            int r = findRoot(v);

            /*
                The current tree node of this component becomes
                a child of the new internal node.
            */
            children[cur].push_back(nodeOf[r]);

            /*
                Merge this component into base.
            */
            parent[r] = base;
        }

        /*
            The merged component is now represented by cur.
        */
        base = findRoot(base);
        nodeOf[base] = cur;
    }

    /*
        Compute the smallest leaf/object ID in every subtree.
    */
    vector<int> subtreeMin(totalNodes + 1, INT_MAX);

    for (int i = 1; i <= N; i++) {
        subtreeMin[i] = i;
    }

    /*
        Internal nodes were created after their children,
        so increasing node ID order is a valid bottom-up order.
    */
    for (int u = N + 1; u <= totalNodes; u++) {
        for (int child : children[u]) {
            subtreeMin[u] = min(subtreeMin[u], subtreeMin[child]);
        }
    }

    /*
        Sort children of every internal node by their smallest leaf.
        This produces the lexicographically smallest order inside each subtree.
    */
    auto bySubtreeMin = [&](int a, int b) {
        return subtreeMin[a] < subtreeMin[b];
    };

    for (int u = N + 1; u <= totalNodes; u++) {
        sort(children[u].begin(), children[u].end(), bySubtreeMin);
    }

    /*
        Collect final forest roots.

        Every remaining DSU root corresponds to one connected component
        of the dendrogram forest.
    */
    vector<int> roots;

    for (int i = 1; i <= N; i++) {
        if (findRoot(i) == i) {
            roots.push_back(nodeOf[i]);
        }
    }

    /*
        Sort forest roots using the same rule.
    */
    sort(roots.begin(), roots.end(), bySubtreeMin);

    /*
        Traverse the sorted forest and produce the final leaf order.

        We use an iterative DFS to avoid recursion depth issues.
    */
    vector<int> order;
    order.reserve(N);

    vector<int> stackNodes;

    /*
        Push in reverse order so that the smallest root is processed first.
    */
    for (auto it = roots.rbegin(); it != roots.rend(); ++it) {
        stackNodes.push_back(*it);
    }

    while (!stackNodes.empty()) {
        int u = stackNodes.back();
        stackNodes.pop_back();

        if (u <= N) {
            /*
                Leaf node: append the object ID.
            */
            order.push_back(u);
        } else {
            /*
                Internal node: process children left-to-right.
                Since this is a stack, push them in reverse order.
            */
            for (auto it = children[u].rbegin(); it != children[u].rend(); ++it) {
                stackNodes.push_back(*it);
            }
        }
    }

    /*
        Answer queries.
        Query positions are 1-indexed.
    */
    for (int pos : queries) {
        cout << order[pos - 1] << '\n';
    }

    return 0;
}
```

---

## 5. Python Implementation with Detailed Comments

```python
import sys


def solve():
    data = list(map(int, sys.stdin.buffer.read().split()))
    idx = 0

    # Read N, M, Q
    n = data[idx]
    idx += 1

    m = data[idx]
    idx += 1

    q = data[idx]
    idx += 1

    # Read horizontal lines
    lines = []

    for _ in range(m):
        y = data[idx]
        idx += 1

        length = data[idx]
        idx += 1

        members = data[idx:idx + length]
        idx += length

        lines.append((y, members))

    # Read queries
    queries = data[idx:idx + q]

    # Larger Y means lower in the dendrogram, so process in decreasing Y.
    lines.sort(key=lambda item: -item[0])

    total_nodes = n + m

    # DSU parent array
    parent = list(range(n + 1))

    # node_of[root] is the current tree node representing this DSU component.
    node_of = list(range(n + 1))

    # children[u] stores children of tree node u.
    children = [[] for _ in range(total_nodes + 1)]

    def find(x):
        """
        Iterative DSU find with path compression.
        """
        while parent[x] != x:
            parent[x] = parent[parent[x]]
            x = parent[x]
        return x

    next_node_id = n

    # Build the forest.
    for y, members in lines:
        next_node_id += 1
        cur = next_node_id

        # Use the first member's component as the base.
        base = find(members[0])

        for v in members:
            r = find(v)

            # The current tree node of this component becomes a child.
            children[cur].append(node_of[r])

            # Merge this component into base.
            parent[r] = base

        # The merged component is represented by the new internal node.
        base = find(base)
        node_of[base] = cur

    # subtree_min[u] = smallest leaf/object ID inside subtree u.
    subtree_min = [10**18] * (total_nodes + 1)

    # Leaves contain themselves.
    for i in range(1, n + 1):
        subtree_min[i] = i

    # Internal nodes were created after their children.
    # Therefore increasing node ID order works.
    for u in range(n + 1, total_nodes + 1):
        best = 10**18

        for child in children[u]:
            if subtree_min[child] < best:
                best = subtree_min[child]

        subtree_min[u] = best

    # Sort children of each internal node by smallest leaf in that child subtree.
    for u in range(n + 1, total_nodes + 1):
        children[u].sort(key=lambda node: subtree_min[node])

    # Collect forest roots.
    roots = []

    for i in range(1, n + 1):
        if find(i) == i:
            roots.append(node_of[i])

    # Sort roots using the same lexicographic rule.
    roots.sort(key=lambda node: subtree_min[node])

    # Traverse sorted forest to get final leaf order.
    order = []

    # Stack-based DFS.
    # Push roots in reverse order so popping processes them left-to-right.
    stack = roots[::-1]

    while stack:
        u = stack.pop()

        if u <= n:
            # Leaf node
            order.append(u)
        else:
            # Internal node.
            # Push children in reverse order because this is a stack.
            for child in reversed(children[u]):
                stack.append(child)

    # Answer queries.
    output = []

    for pos in queries:
        output.append(str(order[pos - 1]))

    sys.stdout.write("\n".join(output))


if __name__ == "__main__":
    solve()
```