// failed/solution-prabowo-fail.cpp
#include "islands.h"
#include <functional>
#include <queue>
#include <utility>
#include <variant>
#include <vector>
std::vector<std::vector<std::pair<int, int>>> pruneNoOutgoing(
int N, int M, std::vector<int> &U, std::vector<int> &V) {
std::vector<int> deg(N);
std::vector<std::vector<std::pair<int, int>>> radj(N);
for (int i = 0; i < M; ++i) {
if (U[i] < 0) {
continue;
}
radj[V[i]].emplace_back(U[i], i);
++deg[U[i]];
}
std::queue<int> q;
for (int i = 0; i < N; ++i) {
if (deg[i] == 0) {
q.push(i);
}
}
std::vector<bool> removed(N, false);
while (!q.empty()) {
int u = q.front();
q.pop();
removed[u] = true;
for (auto [v, i] : radj[u]) {
if (--deg[v] == 0) {
q.push(v);
}
}
}
std::vector<std::vector<std::pair<int, int>>> adj(N);
for (int i = 0; i < M; ++i) {
if (U[i] < 0 || removed[U[i]] || removed[V[i]]) {
U[i] = V[i] = -1;
} else {
adj[U[i]].emplace_back(V[i], i);
}
}
return adj;
}
std::variant<bool, std::vector<int>> find_journey(
int N, int M, std::vector<int> U, std::vector<int> V) {
std::vector<std::vector<std::pair<int, int>>> adj = pruneNoOutgoing(N, M, U, V);
std::vector<int> inverted(M);
int invertedCnt = 0;
std::vector<int> ans;
int lstEdge = -1;
std::function<int(int)> step = [&](int edge) {
int u = U[edge], v = V[edge];
if (lstEdge != -1) {
adj[U[lstEdge]].emplace_back(V[lstEdge], lstEdge);
}
lstEdge = edge;
for (int i = 0; i < static_cast<int>(adj[u].size()); ++i) {
if (adj[u][i].second == edge) {
adj[u][i] = adj[u].back();
adj[u].pop_back();
break;
}
}
std::swap(U[lstEdge], V[lstEdge]);
if (inverted[lstEdge] ^= 1) {
++invertedCnt;
} else {
--invertedCnt;
}
ans.push_back(edge);
return v;
};
// The WA part is because the next section is not made into a loop
// I only step out of the 1-outdegree and prune once
// And even if it was made into a loop, the complexity would be O(N^2)
// WA case is very easy: a bidirectional line graph of size > 3
int u = 0;
while (static_cast<int>(adj[u].size()) == 1) {
u = step(adj[u][0].second);
}
std::vector<int> pathToTwoDeg = ans;
for (int j : ans) {
U[j] = V[j] = -1;
}
adj = pruneNoOutgoing(N, M, U, V);
if (static_cast<int>(adj[u].size()) == 0) {
return false;
}
invertedCnt = 0;
lstEdge = -1;
do {
u = step(adj[u].back().second);
} while (invertedCnt != 0);
while (!pathToTwoDeg.empty()) {
ans.push_back(pathToTwoDeg.back());
pathToTwoDeg.pop_back();
}
return ans;
}
