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longest-special-path-ii.cpp
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// Time: O(n + e)
// Space: O(n + e)
// iterative dfs, two pointers, sliding window, prefix sum
class Solution {
public:
vector<int> longestSpecialPath(vector<vector<int>>& edges, vector<int>& nums) {
vector<vector<pair<int, int>>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1], e[2]);
adj[e[1]].emplace_back(e[0], e[2]);
}
const auto& iter_dfs = [&]() {
vector<int> result(2, numeric_limits<int>::max());
unordered_map<int, int> lookup;
vector<int> prefix = {0};
vector<tuple<int, int, int, int, vector<int>, int, int>> stk = {{1, 0, -1, 0, {-1, -1}, -1, -1}};
while (!empty(stk)) {
auto [step, u, p, d, left, i, prev_d] = stk.back(); stk.pop_back();
if (step == 1) {
int prev_d;
tie(prev_d, lookup[nums[u] - 1]) = pair(lookup.count(nums[u] - 1) ? lookup[nums[u] - 1] : -1, d);
vector<int> new_left(left);
int curr = prev_d;
for (int i = 0; i < size(new_left); ++i) {
if (curr > new_left[i]) {
swap(curr, new_left[i]);
}
}
result = min(result, vector<int>{-(prefix[(d - 1) + 1] - prefix[new_left[1] + 1]), d - new_left[1]});
stk.emplace_back(4, u, -1, -1, vector<int>(), -1, prev_d);
stk.emplace_back(2, u, p, d, new_left, 0, -1);
} else if (step == 2) {
if (i == size(adj[u])) {
continue;
}
stk.emplace_back(2, u, p, d, left, i + 1, -1);
auto [v, l] = adj[u][i];
if (v == p) {
continue;
}
prefix.emplace_back(prefix.back() + l);
stk.emplace_back(3, -1, -1, -1, vector<int>(), -1, -1);
stk.emplace_back(1, v, u, d+1, left, -1, -1);
} else if (step == 3) {
prefix.pop_back();
} else if (step == 4) {
lookup[nums[u] - 1] = prev_d;
}
}
return vector<int>{-result[0], result[1]};
};
return iter_dfs();
}
};
// Time: O(n + e)
// Space: O(n + e)
// dfs, two pointers, sliding window, prefix sum
class Solution2 {
public:
vector<int> longestSpecialPath(vector<vector<int>>& edges, vector<int>& nums) {
vector<vector<pair<int, int>>> adj(size(nums));
for (const auto& e : edges) {
adj[e[0]].emplace_back(e[1], e[2]);
adj[e[1]].emplace_back(e[0], e[2]);
}
unordered_map<int, int> lookup;
vector<int> prefix = {0};
vector<int> result(2, numeric_limits<int>::max());
const function<void (int, int, int, const vector<int>&)> dfs = [&](int u, int p, int d, const vector<int>& left) {
int prev_d;
tie(prev_d, lookup[nums[u] - 1]) = pair(lookup.count(nums[u] - 1) ? lookup[nums[u] - 1] : -1, d);
vector<int> new_left(left);
int curr = prev_d;
for (int i = 0; i < size(new_left); ++i) {
if (curr > new_left[i]) {
swap(curr, new_left[i]);
}
}
result = min(result, vector<int>{-(prefix[(d - 1) + 1] - prefix[new_left[1] + 1]), d - new_left[1]});
for (const auto& [v, l] : adj[u]) {
if (v == p) {
continue;
}
prefix.emplace_back(prefix.back() + l);
dfs(v, u, d + 1, new_left);
prefix.pop_back();
}
lookup[nums[u] - 1] = prev_d;
};
dfs(0, -1, 0, {-1, -1});
return {-result[0], result[1]};
}
};