使用邻接矩阵存储加权图，无穷大使用常数MAXLEN代表，然后使用Dijkstra方法求取最短路径

 1 #include <stdio.h>
 2 
 3 #define MAXLEN 1000
 4 int cost[7][7];
 5 int dist[7];
 6 
 7 void creategraph(int* node,int num)
 8 {
 9     int from;
10     int to;
11     int i;
12     for(i = 0;i < num; i++)
13     {
14         from = node[i*3];
15         to = node[i*3+1];
16         cost[from][to] = node[i*3+2];
17     }
18 }
19 
20 void shortestpath(int begin,int num)
21 {
22     int selected[7];
23     int min;
24     int s;
25     int i,j;
26     
27     for(i = 2;i <= num;i++)
28     {
29         selected[i] = 0;
30         dist[i] = cost[begin][i];
31     }
32     selected[begin] = 1;
33     dist[begin] = 0;
34     printf("顶点1     2     3     4     5     6\n");
35     for(j = 1;j <= num;j++)
36         printf(" %4d ",dist[j]);
37     printf("\n");
38     for(i = 1;i <= num - 1;i++)
39     {
40         min = MAXLEN;
41         for(j = 1;j <= num;j++)
42             if(min > dist[j] && selected[j] == 0)
43             {
44                 s = j;
45                 min = dist[j];
46             }
47         selected[s] = 1;
48         for(j = 1;j <= num;j++)
49         {
50             if(selected[j] == 0 &&
51                 dist[s] + cost[s][j] < dist[j])
52                 dist[j] = dist[s] + cost[s][j];
53             printf(" %4d ",dist[j]);
54         }
55         printf("\n");
56     }
57 }
58 
59 int main()
60 {
61     int node[7][3] = {
62         { 1, 2, 35},
63         { 2, 3, 45},
64         { 2, 4, 30},
65         { 3, 5, 25},
66         { 4, 5, 45},
67         { 4, 6, 130},
68         { 5, 6, 100},
69     };
70     int i,j;
71     
72     for(i = 1;i <= 6;i++)
73         for(j = 1;j <= 6;j++)
74             cost[i][j] = MAXLEN;
75     creategraph(node,7);
76     printf("加权图的邻接矩阵内容：\n");
77     for(i = 1;i <= 6;i++)
78     {
79         for(j = 1;j <= 6;j++)
80             printf(" %4d ",cost[i][j]);
81         printf("\n");
82     }
83     printf("\n从顶点1到各顶点最近距离计算过程：\n");
84     shortestpath(1,6);
85 }