Shortest Path, bellman-ford Algorithm

The Problem description is given with n vertices and a directed graph of m edges (some edge weights may be negative, but there is no negative ring ). Calculate the shortest path (from vertex 1 to vertex n) from vertex 1 to other ). The first line of the input format is two integers n, m. In the next m row, each row has three integers u, v, l, indicating that u to v has an edge with a length of l. There are n-1 lines in the output format. line I indicates the shortest path from Point 1 to point I + 1. Sample input 3 31 2-12 3-13 1 2 sample output-1-2 data size and conventions for 10% of data, n = 2, m = 2. For 30% of the data, n <= 5, m <= 10. For 100% of data, 1 <= n <= 20000,1 <= m <= 200000,-10000 <= l <= 10000, to ensure that all other vertices can be reached from any vertex.

1 package job525; 2 3 import java. util. values; 4 5 class Eage {6 int from; 7 int to; 8 int cost; 9 10 public int getFrom () {11 return from; 12} 13 public void setFrom (int from) {14 this. from = from; 15} 16 public int getTo () {17 return to; 18} 19 public void setTo (int to) {20 this. to = to; 21} 22 public int getCost () {23 return cost; 24} 25 public void setCost (int cost) {26 this. cost = cost; 27} 28 publ Ic Eage (int from, int to, int cost) 29 {30 this. from = from; 31 this. to = to; 32 this. cost = cost; 33} 34} 35 36 public class the shortest path of bellman_ford {37 38 static int m, n, inf = 100000; // n: n vertices; m: m side 39 static int [] d = new int [20000]; 40 static Eage [] eg = new Eage [200000]; 41 42 public static void main (String [] args) {43 expect SC = new expect (System. in); 44 n = SC. nextInt (); m = SC. nextInt (); 45 int I = 0; 46 while (I <m) 47 {48 Int u, v, l; 49 u = SC. nextInt (); 50 v = SC. nextInt (); 51 l = SC. nextInt (); 52 eg [I ++] = new Eage (u, v, l); 53} 54 // test Storage Limit 55/* 56 for (I = 0; I <m; I ++) 57 {58 System. out. println ("(" + eg [I]. getFrom () + "," + eg [I]. getTo () + "):" + eg [I]. getCost (); 59} 60 */61 // test ended -------------------------------------------------------------------------------------- 62 Bellman_ford (); 63 for (I = 2; I <= n; I ++) 64 {65 System. out. println (d [I]); 66} 67} 68 public static void bellman_ford () 69 {70 init (); 71 while (true) 72 {73 boolean flag = false; 74 for (int I = 0; I <m; I ++) 75 {76 Eage e = eg [I]; 77 if (d [e. getFrom ()]! = Inf & d [e. getTo ()]> d [e. getFrom ()] + e. getCost () 78 {79 d [e. getTo ()] = d [e. getFrom ()] + e. getCost (); 80 // System. out. println ("(" + eg [I]. getFrom () + "," + eg [I]. getTo () + "):" + eg [I]. getCost () + d [e. getTo ()]); // test 81 flag = true; 82} 83} 84 if (! Flag) 85 break; 86} 87} 88 public static void init () 89 {90 for (int I = 1; I <= n; I ++) 91 d [I] = inf; 92 d [1] = 0; 93} 94}View Code