Implementation of the shortest Bellman-Ford HDU2544

Implementation of the shortest Bellman-Ford HDU2544
Problem Description

In each year's competition, all the finalists will get a very beautiful t-shirt. However, every time our staff moved hundreds of pieces of clothing from the store back to the stadium, they were very tired! So now they want to find the shortest route from the store to the stadium. Can you help them?

Input

The input includes multiple groups of data. The first row of each group of data is two integers, N and M (N <= 100, M <= 10000). N indicates several intersections on the streets of Chengdu, the intersection marked as 1 is the location of the store, the intersection marked as N is the location of the stadium, and M represents several roads in Chengdu. N = M = 0 indicates that the input is complete. In the next M row, each row contains three integers, A, B, and C (1 <= A, B <= N, 1 <= C <= 1000 ), it means there is A road between Intersection A and intersection B. Our staff need to walk this road in C minutes.
Enter a route to ensure there is at least one store.

Output

Output a line for each group of inputs, indicating the shortest time for a staff member to walk from the store to the stadium

Sample Input

2 1
1 2 3
3 3
1 2 5
2 3 5
3 1 2
0 0

Sample Output

3
2

Solutions

In an undirected graph, Bellman-Ford can expand edges to 2 * e and convert them to a "Directed Graph" to solve the problem.

Code

# Include
  
   
Const int max_v = 110; const int max_e = 10010; const int INF = 1000000000; int e, v; struct edge {int from, to, cost ;}; int d [max_v]; edge eg [max_e * 2]; int main () {while (scanf ("% d", & v, & e) & v) {// the edge is extended to the range of 1-2E for (int I = 0; I <max_v; I ++) d [I] = INF; for (int I = 1; I <= e; I ++) {scanf ("% d", & eg [I]. from, & eg [I]. to, & eg [I]. cost); eg [I + e]. from = eg [I]. to; eg [I + e]. to = eg [I]. from; eg [I + e]. cost = eg [ I]. cost;} d [1] = 0; while (true) {bool flag = false; for (int I = 1; I <= 2 * e; I ++) {if (d [eg [I]. from]! = INF & d [eg [I]. from] + eg [I]. cost <d [eg [I]. to]) {d [eg [I]. to] = d [eg [I]. from] + eg [I]. cost; flag = true ;}} if (! Flag) break;} printf ("% d \ n", d [v]);} return 0 ;}