Graph's shortest path solution four
The bellman-ford algorithm
traverses all edges. The edges have a starting point i and an ending point j. If the shortest distance d[i] from the source point to the starting point has been calculated, compare d[j] and d[i]+cost. If the former is more than If the latter is large, you can update d[j] and
so on, until there is no data to update, so that the shortest distance from the source point to all vertices is calculated.
For the previous code, you can extract the edge set first, so you don’t need to do it every time Scan a two-dimensional array
import java.util.ArrayList;
import java.util.List;
public class 图的最短路问题_优化_边集 {
public static void main(String[] args) {
edges = buildEdges(graph);
int[] shortestPath = shortestPath(0);
Util.print(shortestPath);
}
/**
* 求起点到各顶点的最短距离
* @param s 起点
* @return
*/
private static int[] shortestPath(int s) {
int n = graph.length;
//记录s到各顶点的最短距离
int[] d = new int[n];
for (int i = 0; i < n; i++) {
d[i] = Integer.MAX_VALUE;
}
d[s] = 0;//到自己的距离为0
while (true) {
boolean update = false;
for (Edge<Integer> e : edges) {
if (d[e.getStart()] != Integer.MAX_VALUE && d[e.getEnd()] > d[e.getStart()] + e.getDistance()) {
update = true;
d[e.getEnd()] = d[e.getStart()] + e.getDistance();
}
}
if (!update) break;
}
return d;
}
static List<Edge<Integer>> edges;
static List<Edge<Integer>> buildEdges(int[][] graph) {
int n = graph.length;
List<Edge<Integer>> edges = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
int cost = graph[i][j]; // i,j之间的距离
if (cost > 0) {
// i,j 两点之间有边，起点是i
edges.add(new Edge<>(i, j, cost));
}
}
}
return edges;
}
static int[][] graph = {
{
0, 2, 5, 0, 0, 0, 0},
{
2, 0, 4, 6, 10, 0, 0},
{
5, 4, 0, 2, 0, 0, 0},
{
0, 6, 2, 0, 0, 1, 0},
{
0, 10, 0, 0, 0, 3, 5},
{
0, 0, 0, 1, 3, 0, 9},
{
0, 0, 0, 0, 5, 9, 0}
};
}