/*
JOI 2013 Synchronisation
- In this problem, we essentially want to synchronise data between connected components
in a forest. We need to support 4 types of queries:
- Set all nodes in a connected component to have value x
- Find the value in some node
- Split a connected component
- Join 2 connected components
- 1) How do we sync data without actually storing the computers that have synced?
- Let u and v be connected, where u is the parent of v
- For each node v, store a value info[v] which denotes the amount of info in it
- Let last_sync[v] be the amount of data that was common between u and v
the last time they synced
- Clearly, the new amount of data (newval) will be info[u] + info[v] - last_sync[v]
if we connect u and v
- Simply set info[u] = info[v] = last_sync[v] = newval
- 2) How can we represent connected components?
- Root the tree arbitrarily
- Each connected component must have a single lowest ancestor
- We can do all operations for that connected component on that ancestor
- We can find it in O(log^2 N) with DFS, binary lifting, and a Fenwick tree
- First find the dfs order of the nodes (i.e. find tin and tout)
- Update a node in the Fenwick tree to be +1 if it has an active edge to its parent
- Notice that we can do path queries if we update tin[node] -> +1 and tout[node] -> -1
- Let query(x) be the sum from the Fenwick tree from 1 to x
- u is an ancestor of v iff query(u) == query(v)
- Thus we can use binary lifting to find the lowest ancestor of a component
- 3) How can we process a split/union?
- Let u and v be the edge concerned, where u is the parent of v
- When we union 2 components, just apply 1) but instead of just u, update lowest_ancestor(u)
- When we split a component, v becomes its component's lowest ancestor, so just update info[v]
and last_sync[v] accordingly
- Also, update the Fenwick tree for both events
- tin[node] -> +1, tout[node] -> -1 for unions
- tin[node] -> 0, tout[node] -> 0 for splits
- Complexity: O(N log^2 N)
*/
#include <bits/stdc++.h>
#define FOR(i, x, y) for (int i = x; i < y; i++)
typedef long long ll;
using namespace std;
int n, m, q;
bool active[100001];
vector<int> graph[100001];
pair<int, int> edges[200001];
int info[100001], last_sync[100001];
// DFS order
int timer = 1, tin[100001], tout[100001];
// Binary lifting parents
int anc[100001][20];
void dfs(int node = 1, int parent = 0) {
anc[node][0] = parent;
for (int i = 1; i < 20 && anc[node][i - 1]; i++) {
anc[node][i] = anc[anc[node][i - 1]][i - 1];
}
info[node] = 1;
tin[node] = timer++;
for (int i : graph[node]) if (i != parent) dfs(i, node);
tout[node] = timer;
}
// Fenwick tree
int bit[100001];
void update(int pos, int val) { for (; pos <= n; pos += (pos & (-pos))) bit[pos] += val; }
int query(int pos) {
int ans = 0;
for (; pos; pos -= (pos & (-pos))) ans += bit[pos];
return ans;
}
// Binary lifting
int find_ancestor(int node) {
int lca = node;
for (int i = 19; ~i; i--) {
if (anc[lca][i] && query(tin[anc[lca][i]]) == query(tin[node])) lca = anc[lca][i];
}
return lca;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cin >> n >> m >> q;
FOR(i, 1, n) {
cin >> edges[i].first >> edges[i].second;
graph[edges[i].first].push_back(edges[i].second);
graph[edges[i].second].push_back(edges[i].first);
}
dfs();
FOR(i, 1, n + 1) {
update(tin[i], -1);
update(tout[i], 1);
}
while (m--) {
int x;
cin >> x;
int u = edges[x].first, v = edges[x].second;
if (anc[u][0] == v) swap(u, v);
if (active[x]) {
info[v] = last_sync[v] = info[find_ancestor(u)];
update(tin[v], -1);
update(tout[v], 1);
} else {
info[find_ancestor(u)] += info[v] - last_sync[v];
update(tin[v], 1);
update(tout[v], -1);
}
active[x] = !active[x];
}
while (q--) {
int x;
cin >> x;
cout << info[find_ancestor(x)] << '\n';
}
return 0;
}
