我目前正在通过否定所有边权重并运行 Bellman-Ford 算法来寻找有向非循环正加权图中的最长路径。这很好用。
但是,我想打印使用了哪些节点/边的轨迹。我该怎么做?
该程序将节点数、源、目的地和边权重作为输入。输入在 -1 -1 -1
处停止。我的代码如下:
import java.util.Arrays;
import java.util.Vector;
import java.util.Scanner;
public class BellmanFord {
public static int INF = Integer.MAX_VALUE;
// this class represents an edge between two nodes
static class Edge {
int source; // source node
int destination; // destination node
int weight; // weight of the edge
public Edge() {}; // default constructor
public Edge(int s, int d, int w) { source = s; destination = d; weight = (w*(-1)); }
}
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int inputgood = 1;
int tail;
int head;
int weight;
int count = -1;
Vector<Edge> edges = new Vector<Edge>(); // data structure to hold graph
int nnodes = input.nextInt();
while (inputgood == 1) {
tail = input.nextInt();
head = input.nextInt();
weight = input.nextInt();
if (tail != -1) {
edges.add(new Edge(tail, head, weight));
count++;
}
if (tail == -1)
inputgood = 0;
}
int start = edges.get(0).source;
bellmanFord(edges, nnodes, start);
}
public static void bellmanFord(Vector<Edge> edges, int nnodes, int source) {
// the 'distance' array contains the distances from the main source to all other nodes
int[] distance = new int[nnodes];
// at the start - all distances are initiated to infinity
Arrays.fill(distance, INF);
// the distance from the main source to itself is 0
distance[source] = 0;
// in the next loop we run the relaxation 'nnodes' times to ensure that
// we have found new distances for ALL nodes
for (int i = 0; i < nnodes; ++i)
// relax every edge in 'edges'
for (int j = 0; j < edges.size(); ++j) {
// analyze the current edge (SOURCE == edges.get(j).source, DESTINATION == edges.get(j).destination):
// if the distance to the SOURCE node is equal to INF then there's no shorter path from our main source to DESTINATION through SOURCE
if (distance[edges.get(j).source] == INF) continue;
// newDistance represents the distance from our main source to DESTINATION through SOURCE (i.e. using current edge - 'edges.get(j)')
int newDistance = distance[edges.get(j).source] + edges.get(j).weight;
// if the newDistance is less than previous longest distance from our main source to DESTINATION
// then record that new longest distance from the main source to DESTINATION
if (newDistance < distance[edges.get(j).destination])
distance[edges.get(j).destination] = newDistance;
}
// next loop analyzes the graph for cycles
for (int i = 0; i < edges.size(); ++i)
// 'if (distance[edges.get(i).source] != INF)' means:
// "
// if the distance from the main source node to the DESTINATION node is equal to infinity then there's no path between them
// "
// 'if (distance[edges.get(i).destination] > distance[edges.get(i).source] + edges.get(i).weight)' says that there's a negative edge weight cycle in the graph
if (distance[edges.get(i).source] != INF && distance[edges.get(i).destination] > distance[edges.get(i).source] + edges.get(i).weight) {
System.out.println("Cycles detected!");
return;
}
// this loop outputs the distances from the main source node to all other nodes of the graph
for (int i = 0; i < distance.length; ++i)
if (distance[i] == INF)
System.out.println("There's no path between " + source + " and " + i);
else
System.out.println("The Longest distance between nodes " + source + " and " + i + " is " + distance[i]);
}
}
最佳答案
您需要稍微修改一下您在 Bellman Ford 实现中所做的事情:
...
int[] lastNode = new int[nnodes];
lastNode[source] = source;
for (int i = 0; i < nnodes; ++i)
for (int j = 0; j < edges.size(); ++j) {
if (distance[edges.get(j).source] == INF) continue;
int newDistance = distance[edges.get(j).source] + edges.get(j).weight;
if (newDistance < distance[edges.get(j).destination])
{
distance[edges.get(j).destination] = newDistance;
lastNode[edges.get(j).destination] = edges.get(j).source;
}
}
然后打印单个路径变为:
static void printPath(int source, int end, int[] lastNodes)
{
if(source!=end)
printPath(source, lastNodes[end], lastNodes);
System.out.print(end+" ");
}
它按从源节点到结束节点的顺序打印路径。
关于java - 使用负边权重的 Bellman-Ford 追踪最长路径,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1776319/