#include "simurgh.h"
#include <bits/stdc++.h>
using namespace std;
// Global state variables
static int edge_status[30005];
static vector<pair<int, int>> adj[505];
static int dpt[505], par_n[505], par_e[505];
static bool vis[505];
static vector<int> tree_edges;
static bool is_tree_edge[30005];
struct DSU {
vector<int> p;
DSU(int n) { p.resize(n); iota(p.begin(), p.end(), 0); }
int find(int i) { return (p[i] == i) ? i : (p[i] = find(p[i])); }
bool unite(int i, int j) {
int root_i = find(i), root_j = find(j);
if (root_i != root_j) { p[root_i] = root_j; return true; }
return false;
}
};
void dfs(int u, int p, int d) {
vis[u] = true;
dpt[u] = d;
for (auto &edge : adj[u]) {
int v = edge.first, id = edge.second;
if (v == p) continue;
if (!vis[v]) {
par_n[v] = u;
par_e[v] = id;
tree_edges.push_back(id);
is_tree_edge[id] = true;
dfs(v, u, d + 1);
}
}
}
vector<int> find_roads(int n, vector<int> u, vector<int> v) {
int m = u.size();
// 1. Reset all structures
tree_edges.clear();
for (int i = 0; i < n; i++) {
adj[i].clear();
vis[i] = false;
}
for (int i = 0; i < m; i++) {
edge_status[i] = -1;
is_tree_edge[i] = false;
}
// 2. Build initial DFS Tree
dfs(0, -1, 0);
// 3. Phase 1: Use cycles to classify tree edges
for (int i = 0; i < m; i++) {
if (is_tree_edge[i] || edge_status[i] != -1) continue;
vector<int> cycle;
int x = u[i], y = v[i];
while (x != y) {
if (dpt[x] > dpt[y]) { cycle.push_back(par_e[x]); x = par_n[x]; }
else { cycle.push_back(par_e[y]); y = par_n[y]; }
}
cycle.push_back(i); // back-edge is the last element
int ref_edge = -1;
for (int e : cycle) {
if (edge_status[e] != -1) {
ref_edge = e;
break;
}
}
vector<int> res(cycle.size(), -1);
int mx = -1, mn = 1e9;
for (int j = 0; j < (int)cycle.size(); j++) {
int cur = cycle[j];
// If we already know the status, we only query the reference edge
if (ref_edge != -1 && edge_status[cur] != -1 && cur != ref_edge) continue;
vector<int> q;
// Build a spanning tree by removing cycle[j] and ensuring i (back-edge) connects it
for (int te : tree_edges) {
bool in_c = false;
for (int ce : cycle) if (ce == te) in_c = true;
if (!in_c) q.push_back(te);
}
for (int k = 0; k < (int)cycle.size() - 1; k++) {
if (k != j) q.push_back(cycle[k]);
}
if (j != (int)cycle.size() - 1) q.push_back(i);
res[j] = count_common_roads(q);
mx = max(mx, res[j]);
mn = min(mn, res[j]);
}
for (int j = 0; j < (int)cycle.size(); j++) {
if (res[j] == -1) continue;
if (mx == mn) {
// If all results are the same, they share status with reference (or are 0)
edge_status[cycle[j]] = (ref_edge != -1 ? edge_status[ref_edge] : 0);
} else {
// The edge whose removal yields the minimum result is the Royal Road (1)
edge_status[cycle[j]] = (res[j] == mn ? 1 : 0);
}
}
}
// Tree edges not part of any cycle are bridges (always royal)
for (int te : tree_edges) if (edge_status[te] == -1) edge_status[te] = 1;
// 4. Phase 2: Binary search for remaining non-tree edges
for (int i = 0; i < n; i++) {
vector<int> candidates;
for (auto &edge : adj[i]) {
int id = edge.second;
int other = (u[id] == i ? v[id] : u[id]);
if (i < other && edge_status[id] == -1) candidates.push_back(id);
}
auto solve = [&](auto self, vector<int> c) -> void {
if (c.empty()) return;
DSU dsu(n);
vector<int> q;
// Add candidates but prevent internal cycles
for (int id : c) if (dsu.unite(u[id], v[id])) q.push_back(id);
int tree_royals_count = 0;
for (int te : tree_edges) {
if (dsu.unite(u[te], v[te])) {
q.push_back(te);
if (edge_status[te] == 1) tree_royals_count++;
}
}
int found_in_set = count_common_roads(q) - tree_royals_count;
if (found_in_set == 0) {
for (int id : c) edge_status[id] = 0;
} else if (found_in_set == (int)c.size()) {
for (int id : c) edge_status[id] = 1;
} else {
int mid = c.size() / 2;
self(self, vector<int>(c.begin(), c.begin() + mid));
self(self, vector<int>(c.begin() + mid, c.end()));
}
};
solve(solve, candidates);
}
vector<int> result;
for (int i = 0; i < m; i++) if (edge_status[i] == 1) result.push_back(i);
return result;
}
