如何理解Dijkstra最短路径算法程序中的细节

Question

我在读Dijkstra最短路径算法。

给出图中的一个图和一个源顶点，求出从源到给定图中所有顶点的最短路径值。

我不明白为什么需要dist[u] != INT_MAX。

#include <stdio.h>
#include <limits.h>

// Number of vertices in the graph
#define V 9

// A utility function to find the vertex with minimum distance value, from
// the set of vertices not yet included in shortest path tree
int minDistance(int dist[], bool sptSet[])
{
   // Initialize min value
   int min = INT_MAX, min_index;

   for (int v = 0; v < V; v++)
     if (sptSet[v] == false && dist[v] <= min)
         min = dist[v], min_index = v;

   return min_index;
}

// A utility function to print the constructed distance array
void printSolution(int dist[], int n)
{
   printf("Vertex   Distance from Source\n");
   for (int i = 0; i < V; i++)
      printf("%d \t\t %d\n", i, dist[i]);
}

// Funtion that implements Dijkstra's single source shortest path algorithm
// for a graph represented using adjacency matrix representation
void dijkstra(int graph[V][V], int src)
{
     int dist[V];     // The output array.  dist[i] will hold the shortest
                      // distance from src to i

     bool sptSet[V]; // sptSet[i] will true if vertex i is included in shortest
                     // path tree or shortest distance from src to i is finalized

     // Initialize all distances as INFINITE and stpSet[] as false
     for (int i = 0; i < V; i++)
        dist[i] = INT_MAX, sptSet[i] = false;

     // Distance of source vertex from itself is always 0
     dist[src] = 0;

     // Find shortest path for all vertices
     for (int count = 0; count < V-1; count++)
     {
       // Pick the minimum distance vertex from the set of vertices not
       // yet processed. u is always equal to src in first iteration.
       int u = minDistance(dist, sptSet);

       // Mark the picked vertex as processed
       sptSet[u] = true;

       // Update dist value of the adjacent vertices of the picked vertex.
       for (int v = 0; v < V; v++)

         // Update dist[v] only if is not in sptSet, there is an edge from 
         // u to v, and total weight of path from src to  v through u is 
         // smaller than current value of dist[v]

         /*  
          * 
          * Why dist[u] != INT_MAX is needed? Can this condition be deleted? 
          *
          */
         if (!sptSet[v] && graph[u][v] && dist[u] != INT_MAX 
                                       && dist[u]+graph[u][v] < dist[v])
            dist[v] = dist[u] + graph[u][v];
     }

     // print the constructed distance array
     printSolution(dist, V);
}

int main(void)
{
   /* Let us create the example graph discussed above */
   int graph[V][V] = {{0, 4, 0, 0, 0, 0, 0, 8, 0},
                      {4, 0, 8, 0, 0, 0, 0, 11, 0},
                      {0, 8, 0, 7, 0, 4, 0, 0, 2},
                      {0, 0, 7, 0, 9, 14, 0, 0, 0},
                      {0, 0, 0, 9, 0, 10, 0, 0, 0},
                      {0, 0, 4, 0, 10, 0, 2, 0, 0},
                      {0, 0, 0, 14, 0, 2, 0, 1, 6},
                      {8, 11, 0, 0, 0, 0, 1, 0, 7},
                      {0, 0, 2, 0, 0, 0, 6, 7, 0}
                     };

    dijkstra(graph, 0);

    return 0;
}

Answer 1

如果没有检查dist[u] != INT_MAX，dist[u]+graph[u][v]可能会导致有符号整数溢出，从而调用未定义的行为。

通过检查，由于短路评估，当dist[u]中的值为INT_MAX时，将不会对INT_MAX进行评估，因此，如果输入成本足够小而不导致溢出，则可以避免未定义的行为。