Problem link: UVa 539 - The Settlers of Catan
(UVa website occasionally becomes down. Alternatively, you can read the problem statement pdf here: UVa External 539 - The Settlers of Catan)
This problem gives a small general graph and asks to find the longest road. The longest road is defined as the longest path within the network that doesn’t use an edge twice. Nodes may be visited more than once, though.
Input Constraints:
1. Number of nodes: n (2 <= n <= 25)
2. Number of edges m (1 <= m <= 25)
3. Edges are un-directed.
4. Nodes have degrees of three or less.
5. The network is not necessarily connected.
Input is given in the format:
15 16
0 2
1 2
2 3
3 4
3 5
4 6
5 7
6 8
7 8
7 9
8 10
9 11
10 12
11 12
10 13
12 14
The first two lines gives the number of nodes n and the number of edges m for this test case respectively. The next m lines describe the m edges. Each edge is given by the numbers of the two nodes connected by it. Nodes are numbered from 0 to n - 1.
The above test can be visualized by the following picture:
Now I know that finding the longest path in a general graph is NP-hard. But as the number of nodes and edges in this problem is small and there's a degree bound of each node, a brute force solution (recursive backtracking) will be able to find the longest path in the given time limit (3.0 seconds).
My strategy to solve the problem was the following:
1. Run DFS (Depth First Search) from each node as the graph can be disconnected
2. When a node visits its neighbor, and that neighbor visits its neighbor and so on, mark the edges as used so that no edge can be used twice in the process
3. When the DFS routine starts to come back to the node from where it began, mark the edges as unused in the unrolling process
4. In each step, update the longest path length
My implementation in C++:
#include <iostream>
#include <vector>
// this function adds an edge to adjacency matrix
// we use this function to build the graph
void addEdgeToGraph(std::vector<std::vector<int>> &graph, int a, int b){
graph[a].emplace_back(b);
graph[b].emplace_back(a); // undirected graph
}
// returns true if the edge between a and b has already been used
bool isEdgeUsed(int a, int b, const std::vector<std::vector<char>> &edges){
return edges[a][b] == '1' || edges[b][a] == '1'; // undirected graph, (a,b) and (b,a) are both valid edges
}
// this function incrementally marks edges when "dfs" routine is called recursively
void markEdgeAsUsed(int a, int b, std::vector<std::vector<char>> &edges){
edges[a][b] = '1';
edges[b][a] = '1'; // order doesn't matter, the edge can be taken in any order [(a,b) or (b,a)]
}
// this function removes edge when a node has processed all its neighbors
// this lets us to reuse this edge in the future to find newer (and perhaps longer) paths
void unmarkEdge(int a, int b, std::vector<std::vector<char>> &edges){
edges[a][b] = '0';
edges[b][a] = '0';
}
int dfs(const std::vector<std::vector<int>> &graph, std::vector<std::vector<char>> &edges, int current_node, int current_length = 0){
int pathLength = -1;
for(int i = 0 ; i < graph[current_node].size() ; ++i){
int neighbor = graph[current_node][i];
if(!isEdgeUsed(current_node, neighbor, edges)){
markEdgeAsUsed(current_node, neighbor, edges);
int ret = dfs(graph, edges, neighbor, current_length + 1);
pathLength = std::max(pathLength, ret);
unmarkEdge(current_node, neighbor, edges);
}
}
return std::max(pathLength, current_length);
}
int dfsFull(const std::vector<std::vector<int>> &graph){
int longest_path = -1;
for(int node = 0 ; node < graph.size() ; ++node){
std::vector<std::vector<char>> edges(graph.size(), std::vector<char>(graph.size(), '0'));
int pathLength = dfs(graph, edges, node);
longest_path = std::max(longest_path, pathLength);
}
return longest_path;
}
int main(int argc, char const *argv[])
{
int n,m;
while(std::cin >> n >> m){
if(!n && !m) break;
std::vector<std::vector<int>> graph(n);
for(int i = 0 ; i < m ; ++i){
int a,b;
std::cin >> a >> b;
addEdgeToGraph(graph, a, b);
}
std::cout << dfsFull(graph) << '\n';
}
return 0;
}
I was ordering what is the worst case for this problem? (I'm wondering it should be n = 25 and m = 25) and in the worst case in total how many times the edges will be traversed? For example for the following test case with 3 nodes and 2 edges:
3 2
0 1
1 2
The dfs routine will be called 3 times, and each time 2 edges will be visited. So in total the edges will be visited 2 x 3 = 6 times. Is there any way to find the upper bound of total edge traversal in the worst case?
