PKU 3259 wormholes-judge the negative weight loop Bellman-Ford

Question:

There are n fields (n <= 500) on the farm, and m paths (M <= 2500) can take t units of time to arrive at J from I. There are also W wormhole holes (W <= 200), which can arrive at the other J from the ground I and timeRegressingT! Note that the path is bidirectional and the wormhole is unidirectional. Now Farmer John wants to know if he can start from a certain place and return to this place so that he can return earlier than the departure time (he can meet himself ^ ).

Analysis:

The disguise of the question is removed. It can be seen that the question requires a directed graph with negative weights.Whether a loop with a negative weight exists.

This problem isBellman-FordThe problem to be solved when finding the shortest path with a negative weight directed graph.

 

1. /*
2. Pku3259 wormholes
3. */
5. # Include <stdio. h>
6. # Include <memory. h>
8. # Define CLR (a) memset (A, 0, sizeof ())
9. # Define Inf (1 <28)
10. # Define n 505
11. # Define M 6005
13. Int bellman_ford (int e [] [3], int M, int N, int d [])
14. {
15. Int K, V, flag, I, J;
16. For (k = 0; k <n; k ++) d [k] = inf;
17. // D [0] = 0;
18. For (k = 0; k <n-1; k ++)
19. {
20. Flag = 0;
21. For (V = 0; v <m; V ++ ){
22. I = E [v] [0]; j = E [v] [1];
23. If (d [J]> d [I] + E [v] [2]) {
24. D [J] = d [I] + E [v] [2];
25. Flag = 1;
26. }
27. }
28. If (! Flag) return 1;
29. }
30. For (v = 0; v <m; v ++)
31. If (d [e [v] [1]> d [e [v] [0] + e [v] [2])
32. Return 0;
33. Return 1;
34. }
36. Int n, m, w;
37. Int e [M] [3], ne;
38. Int d [N];
40. Void addEdge (int I, int j, int k ){
41. E [ne] [0] = I;
42. E [ne] [1] = j;
43. E [ne] [2] = k;
44. Ne ++;
45. }
47. Int main ()
48. {
49. Int I, j, k, t, T;
51. Scanf ("% d", & T );
52. While (T --){
53. // Input
54. Ne = 0;
55. Scanf ("% d", & n, & m, & w );
56. For (k = 0; k <m + w; k ++ ){
57. Scanf ("% d", & I, & j, & t );
58. If (k <m ){
59. AddEdge (I, j, t );
60. AddEdge (j, I, t );
61. }
62. Else {
63. AddEdge (I, j,-t );
64. }
65. }
66. // Output
67. If (Bellman_Ford (e, ne, n, d) puts ("NO ");
68. Else puts ("YES ");
69. }
71. Return 0;
72. }