单元最短路径

问题分析

  • 可参考图的应用——最短路径

Java 源代码

内含详细注释

package Dijkstra;

public class Dijkstra {

  public static void main(String[] args) {
    float max = Float.MAX_VALUE;
    float[][] a = new float[][] {{0, 0, 0, 0, 0, 0},
              {0, max, 10, max, 30, 100},
              {0, max, max, 60, max, max},
              {0, max, max, max, max, 20},
              {0, max, max, 10, max, 50},
              {0, max, max, max, max, max}};
    int v = 1;
    int n = 5;
    float[] dist = new float[n + 1];
    int[] prev = new int[n + 1];
    dijkstra(v, a, dist, prev);
    for (int i = 1; i <= n; i++) {
      if (i != v) {
        System.out.println(v + "->" + i + " : " + dist[i]);
      }
    }
  }
  
  /**
   * 单元最短路径问题的 Dijkstra 算法 
   * @param v: 源顶点
   * @param a：邻接矩阵
   * @param dist：存放最短路径
   * @param prev：存放当前顶点的前一个顶点
   */
  public static void dijkstra(int v, float[][] a, float[] dist, int[] prev) {
    
    int n = dist.length - 1;
    if (v < 1 || v > n) return;
    // 代表当前顶点是否被添加到集合S中
    boolean[] s = new boolean[n + 1];
    // 初始化
    for (int i = 1; i <= n; i++) {
      dist[i] = a[v][i];
      s[i] = false;
      if (dist[i] == Float.MAX_VALUE) prev[i] = 0;
      else prev[i] = v;
    }
    dist[v] = 0;
    s[v] = true;
    for (int i = 1; i < n; i++) {
      float temp = Float.MAX_VALUE;
      int u = v;
      for (int j = 1; j <= n; j++) {  // 寻找不在S中的最小值
        if (!s[j] && (dist[j] < temp)) {  // 从剩余顶点选择
          u = j;
          temp = dist[j];
        }
      }
      // 加入s中
      s[u] = true;
      for (int j = 1; j <= n; j++) {  // 更新dist
        if ((!s[j]) && (a[u][j] < Float.MAX_VALUE)){
          float newdist = dist[u] + a[u][j];
          if (newdist < dist[j]) {
            dist[j] = newdist;
            prev[j] = u;
          }
        }
      }
    }
  }
}
1->2 : 10.0
1->3 : 40.0
1->4 : 30.0
1->5 : 60.0