本博客只做代碼訓練， 理論閱讀請直接點

https://www.cnblogs.com/ECJTUACM-873284962/p/6995648.html

解決圖中最短路徑問題

一、解法

核心思想是DP。

• i頂點
• j目標點
• k經過的點

狀態轉移矩陣：f[k][i][j] = min(f[k-1][i][j], f[k-1][i][k]+f[k-1][k][j])

 1 package algorithm;
 2 
 3 public class Floyd {
 4     private int n;
 5 
 6     private class Graph {
 7         int
 
[][] edges;
 8         char[] vertex;
 9     }
10 
11     void floyd(Graph g) {
12         int[][] A = new int[n][n];
13         for (int i = 0; i < n; i++) {
14             for (int j = 0; j < n; j++) {
15                 A[i][j] = g.edges[i][j];
16             }
17         }
18 
19         for (int
 
 k = 0; k < n; k++) {
20             for (int i = 0; i < n; i++)
21                 for (int j = 0; j < n; j++) {
22                     A[i][j] = Math.min(A[i][j], A[i][k] + A[k][j]);
23                 }
24         }
25     }
26 }

（二）Floyd算法