我的大学有一项任务，已经成功实施了 Dijkstra 和 Bellman-Ford，但我在这方面遇到了麻烦。一切看起来都很好，但它并没有给我正确的答案。
这是代码：
void FloydWarshall()
{
//Also assume that n is the number of vertices and edgeCost(i,i) = 0
int path[500][500];
/* A 2-dimensional matrix. At each step in the algorithm, path[i][j] is the shortest path
from i to j using intermediate vertices (1..k−1). Each path[i][j] is initialized to
edgeCost(i,j) or infinity if there is no edge between i and j.
*/
for(int i = 0 ; i <= nvertices ; i++)
for(int j = 0 ; j <= nvertices ; j++)
path[i][j] = INFINITY;
for(int j = 0 ; j < narestas ; j++) //narestas = number of edges
{
path[arestas[j]->v1][arestas[j]->v2] = arestas[j]->peso; //peso = weight of the edge (aresta = edge)
path[arestas[j]->v2][arestas[j]->v1] = arestas[j]->peso;
}
for(int i = 0 ; i <= nvertices ; i++) //path(i, i) = 0
path[i][i] = 0;
//test print, it's working fine
//printf("\n\n\nResultado FloydWarshall:\n");
//for(int i = 1 ; i <= nvertices ; i++)
// printf("distancia ao vertice %d: %d\n", i, path[1][i]);
// Here's the problem, it messes up, and even a edge who costs 4, and the minimum is 4, it prints 2.
//for k = 1 to n
for(int k = 1 ; k <= nvertices ; k++)
//for i = 1 to n
for(int i = 1 ; i <= nvertices ; i++)
//for j := 1 to n
for(int j = 1 ; j <= nvertices ; j++)
if(path[i][j] > path[i][k] + path[k][j])
path[i][j] = path[i][k] + path[k][j];
printf("\n\n\nResultado FloydWarshall:\n");
for(int i = 1 ; i <= nvertices ; i++)
printf("distancia ao vertice %d: %d\n", i, path[1][i]);
}
我正在使用我制作的这个图表示例：
6 7
1 2 4
1 5 1
2 3 1
2 5 2
5 6 3
6 4 6
3 4 2
意味着我们有 6 个顶点（1 到 6）和 7 个边（1,2），权重为 4... 等等。
如果有人需要更多信息，我愿意提供，只是厌倦了查看此代码而没有发现错误。