我需要创建一个测试,如果图(有向图)作为参数具有负权重的循环,则返回true,否则返回false。
现在我创建了这个。理论上应该检查是否存在"一般"循环,而不是是否存在负循环。我怎样才能改变方法?有更简单的还是更有效的?
//if there is a negative cycle, get out and return
public void bellmanFord(Graph<V, E> graph, V source, V dest) {
ArrayList<V> vertices = (ArrayList<V>) graph.getVertices();
HashMap<V, Boolean> visited = new HashMap<V, Boolean>(vertices.size());
for(V v : vertices) {
visited.put(v, false);
}
boolean cycle = hasNegativeCycle(graph, source, visited, vertices);
if(cycle == true)
return;
else {
...
}
}
public boolean hasNegativeCycle(Graph<V, E> graph, V source, HashMap<V, Boolean> visited, ArrayList<V> vertices) {
visited.put(source, true);
for(V u : vertices) {
ArrayList<V> neigh_u = (ArrayList<V>) graph.getNeighbors(u);
for(V v : neigh_u) {
if(visited.get(v) == true || hasNegativeCycle(graph, v, visited, vertices)) {
return true;
}
}
}
return false;
}
感谢编辑:正如你可以从上面写的方法名看到的,我正试图实现Bellman-Ford算法,我遵循这个伪代码:
BellmanFord(Graph G, Vertex start) {
foreach(Vertex u of G) {
dist[u] = ∞;
prev[u] = -1;
}
prev[start] = s;
dist[start] = 0;
repeat n times {
foreach(Vertex u of G) {
foreach(Vertex v near u) {
if(dist[u] + weigth_uv < dist[v]) {
prev[v] = u;
dist[v] = dist[u] + weigth_uv;
}
}
}
}
}
必须应用Bellman-Ford算法。Wikipedia
具有适当的伪代码。如果你正确地使用这个方法,你的问题就会解决。
function BellmanFord(list vertices, list edges, vertex source)
::distance[],predecessor[]
// This implementation takes in a graph, represented as
// lists of vertices and edges, and fills two arrays
// (distance and predecessor) with shortest-path
// (less cost/distance/metric) information
// Step 1: initialize graph
for each vertex v in vertices:
if v is source then distance[v] := 0
else distance[v] := inf
predecessor[v] := null
// Step 2: relax edges repeatedly
for i from 1 to size(vertices)-1:
for each edge (u, v) in Graph with weight w in edges:
if distance[u] + w < distance[v]:
distance[v] := distance[u] + w
predecessor[v] := u
// Step 3: check for negative-weight cycles
for each edge (u, v) in Graph with weight w in edges:
if distance[u] + w < distance[v]:
error "Graph contains a negative-weight cycle"
return distance[], predecessor[]
您可能想要对图进行BFS遍历。在每个节点访问时,记录节点的唯一id(例如,如果实现了。hashcode())到HashSet中。每当您试图将一个已经存在的元素插入哈希集时,您都会发现一个圆圈。
如果你在节点F找到了一个圆圈,你可以通过向上遍历树来计算圆圈的总权重,直到你再次找到F,并将权重相加。
当然,在确定圆的大小之后,如果它是正的,你必须继续BFS遍历,但是不遍历F的子节点。如果它是负的,从函数返回,因为你发现了一个负的圆。
编辑:您还可以在BFS遍历步骤期间跟踪当前的总权重,这样您就不必向上遍历树来计算总权重…当你的图形是有向的,这种方法也会更适合…