/**
* Not too difficult~ Let P_1, P_2, ..., P_k be the connected components of the graph.
* Clearly, every P_i is a tree.
* [Lemma] There exists a way of connecting the edges that minimizes the resulting tree's
* diameter in such a way that,
* for a fixed i, all edges connecting P_i and another P_j (i!=j) are
* incident to the same node in P_i.
* This is a greedy strategy. Greedily, we can again prove that the node in P_i has the
* smallest f(node) := max{ dist(node, n2) : n2 is a node in P_i different from node }.
* Next, we can view each P_i as a "super node" and consider how we shall connect them.
* Since we have to minimize the diameter, we shall connect the super nodes to form a star
* graph, whose diameter is at most 2. Again, we shall be greedy: the center super node in
* our star graph should be the component with the largest f(chosen_node_in_the_comp).
*
* Time Complexity: O(n * log(n)) (Optimal: O(n); O(n * log(n)) is just simpler)
* Implementation 1 (Full solution, greedy, tree dp)
*/
#include <bits/stdc++.h>
#include "dreaming.h"
typedef std::vector<int> vec;
struct edge_t {
int node, len;
};
typedef std::vector<edge_t> adj_list_t;
std::vector<adj_list_t> trees;
vec down, down_node, down2, up;
std::vector<bool> visited;
int diameter = 0; // diameter of the resulting graph
void dfs1(int k, int parent) {
visited[k] = true;
for (const edge_t& e : trees[k]) {
int child = e.node;
if (parent == child)
continue;
dfs1(child, k);
int d = down[child] + e.len;
if (d > down[k])
down2[k] = down[k], down[k] = d, down_node[k] = child;
else
down2[k] = std::max(down2[k], d);
}
}
int dfs2(int k, int parent) {
int cost = std::max(up[k], down[k]);
diameter = std::max(diameter, cost);
for (const edge_t& e : trees[k]) {
int child = e.node;
if (parent == child)
continue;
up[child] = (down_node[k] != child ? down[k] : down2[k]) + e.len;
up[child] = std::max(up[child], up[k] + e.len);
cost = std::min(cost, dfs2(child, k));
dfs2(child, k);
}
return cost;
}
int travelTime(int n, int m, int L, int A[], int B[], int T[]) {
trees.assign(n, adj_list_t());
for (int k = 0; k < m; k++) {
trees[A[k]].push_back(edge_t{B[k], T[k]});
trees[B[k]].push_back(edge_t{A[k], T[k]});
}
visited.assign(n, false);
down.assign(n, 0);
down_node.resize(n);
down2.assign(n, 0);
up.assign(n, 0);
vec costs;
for (int k = 0; k < n; k++) {
if (visited[k])
continue;
dfs1(k, -1);
costs.push_back(dfs2(k, -1));
}
// std::sort(costs.rbegin(), costs.rend());
// while (int(costs.size()) > 3)
// costs.pop_back();
// for (int i = 1; i < int(costs.size()); i++)
// costs[i] += L;
// std::sort(costs.rbegin(), costs.rend());
// diameter = std::max(diameter, costs[0] + (int(costs.size()) > 1 ? costs[1] : 0));
return diameter;
}
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Execution timed out |
1068 ms |
11804 KB |
Time limit exceeded |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
59 ms |
304 KB |
Output is correct |
| 2 |
Correct |
1 ms |
212 KB |
Output is correct |
| 3 |
Execution timed out |
1088 ms |
212 KB |
Time limit exceeded |
| 4 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Execution timed out |
1068 ms |
11804 KB |
Time limit exceeded |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
13 ms |
6788 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Correct |
59 ms |
304 KB |
Output is correct |
| 2 |
Correct |
1 ms |
212 KB |
Output is correct |
| 3 |
Execution timed out |
1088 ms |
212 KB |
Time limit exceeded |
| 4 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Execution timed out |
1068 ms |
11804 KB |
Time limit exceeded |
| 2 |
Halted |
0 ms |
0 KB |
- |
