Poj 3259 wormholes (Bellman-Ford)

Link: poj 3259

A famer has some farms, some of which contain some fields, some wormhole holes in the field, and a path (two-way) between the field and the field ), that is, the time from A to B and from B to a is C. the nature of wormhole: Time backflow. That is, the time spent from A to B through the wormhole is-C (one-way). I asked if he could go back to the starting point through the nature of the wormhole.

Idea: This question is actually to determine whether there is a negative weight loop. You can use the spfa algorithm or the Bellman-Ford algorithm to determine whether there is a negative weight loop. If a negative weight loop exists, this can be achieved; otherwise, it cannot.

Bellman-Ford Algorithm

# Include <stdio. h> # include <limits. h> struct Stu {int A, B, C;} edge [6000]; int dis [550], n, m; int bellmanford () {int flag = 0, I, j; for (I = 1; I <= N; I ++) dis [I] = int_max; DIS [1] = 0; for (I = 1; I <= n-1; I ++) for (j = 1; j <= m; j ++) if (DIS [edge [J]. a]! = Int_max & dis [edge [J]. a] + edge [J]. c <dis [edge [J]. b]) dis [edge [J]. b] = dis [edge [J]. a] + edge [J]. c; for (I = 1; I <= m; I ++) // determine the negative weight loop if (DIS [edge [I]. a]! = Int_max & dis [edge [I]. a] + edge [I]. c <dis [edge [I]. b]) {flag = 1; break;} return flag;} int main () {int T, I, J, K, flag; scanf ("% d ", & T); While (t --) {scanf ("% d", & N, & M, & K); j = 1; for (I = 1; I <= m; I ++) {scanf ("% d", & edge [J]. a, & edge [J]. b, & edge [J]. c); j ++; edge [J]. A = edge [J-1]. b; // The path is a bidirectional edge [J]. B = edge [J-1]. a; edge [J]. C = edge [J-1]. c; j ++ ;}for (I = 1; I <= K; I ++) {scanf ("% d", & edge [J]. a, & edge [J]. b, & edge [J]. c); edge [J]. C =-edge [J]. c; // the wormhole is unidirectional and the time is negative J ++;} M = J-1; flag = bellmanford (); If (FLAG) printf ("Yes \ n "); else printf ("NO \ n");} return 0 ;}