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更新0094.城市间货物运输I.md Java版本解法
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@ -392,6 +392,63 @@ Bellman_ford 是可以计算 负权值的单源最短路算法。
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## 其他语言版本
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### Java
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```Java
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public class Main {
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// Define an inner class Edge
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static class Edge {
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int from;
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int to;
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int val;
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public Edge(int from, int to, int val) {
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this.from = from;
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this.to = to;
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this.val = val;
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}
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}
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public static void main(String[] args) {
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// Input processing
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Scanner sc = new Scanner(System.in);
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int n = sc.nextInt();
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int m = sc.nextInt();
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List<Edge> edges = new ArrayList<>();
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for (int i = 0; i < m; i++) {
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int from = sc.nextInt();
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int to = sc.nextInt();
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int val = sc.nextInt();
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edges.add(new Edge(from, to, val));
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}
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// Represents the minimum distance from the current node to the original node
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int[] minDist = new int[n + 1];
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// Initialize the minDist array
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Arrays.fill(minDist, Integer.MAX_VALUE);
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minDist[1] = 0;
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// Starts the loop to relax all edges n - 1 times to update minDist array
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for (int i = 1; i < n; i++) {
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for (Edge edge : edges) {
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// Updates the minDist array
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if (minDist[edge.from] != Integer.MAX_VALUE && (minDist[edge.from] + edge.val) < minDist[edge.to]) {
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minDist[edge.to] = minDist[edge.from] + edge.val;
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}
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}
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}
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// Outcome printing
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if (minDist[n] == Integer.MAX_VALUE) {
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System.out.println("unconnected");
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} else {
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System.out.println(minDist[n]);
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}
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}
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}
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```
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### Python
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