dijk

.....................用矩阵存..................... 1 int mp[N][N];
 2 bool p[N];
 3 int dist[N];
 4 void dijk(int s , int n)
 5 {
 6     int i , j , k ;
 7     for( i = 1 ; i <= n ;i++)
 8     {
 9         p[i] = false;
10         dist[i] = mp[s][i];
11     }
12     p[s] = true;
13     dist[s] = 0;
14     for(i = 1 ; i < n ; i++)
15     {
16         int Min = INF;
17         int k = 0 ;
18         for( j = 1 ; j <= n ;j++)
19         {
20             if(!p[j]&&dist[j]<Min)
21             {
22                 Min = dist[j];
23                 k = j;
24             }
25         }
26         if(Min==INF) return ;
27         p[k] = true;
28         for(j = 1 ; j <= n ;j++)
29         {
30             if(!p[j]&&mp[k][j]!=INF&&dist[j]>dist[k]+mp[k][j])
31                 dist[j] = dist[k]+mp[k][j];
32         }
33     }
34 }

..............用链表存................. 1 struct Edge{
 2     int to;
 3     int v ;
 4     int next;
 5 }edge[M];
 6 int Enct;
 7 int head[N];
 8 void add(int from ,int to , int v )
 9 {
10     edge[Enct].to = to ;
11     edge[Enct].v = v ;
12     edge[Enct].next = head[from];
13     head[from] = Enct++;
14
15     edge[Enct].to = from ;
16     edge[Enct].v = v ;
17     edge[Enct].next = head[to];
18     head[to] = Enct++;
19 }
20 void init ()
21 {
22     Enct = 0;
23     memset(head,-1, sizeof(head));
24 }
25 int dist[N];
26 bool p[N];
27 void dijk(int s, int n)
28 {
29     int i , j , k ;
30     for( i = 1 ;  i <= n ; i++ )
31     {
32         p[i] = false;
33         dist[i]= INF;
34     }
35     p[s] = true;
36     dist[s] = 0;
37     for( i = head[s] ; i != -1 ; i = edge[i].next)
38     {
39         Edge e = edge[i];
40         if(e.v<dist[e.to])//考虑重边
41         dist[e.to] = e.v;
42     }
43     for( i = 1 ;i < n ;i++)
44     {
45         int Min = INF;
46         int k = 0 ;
47         for( j = 1 ; j <= n ; j++)
48         {
49             if(!p[j]&&dist[j]<Min)
50             {
51                 Min = dist[j];
52                 k = j;
53             }
54         }
55         p[k] = true;
56         if(Min == INF) return ;
57         for( j = head[k] ;j!=-1 ;j=edge[j].next)
58         {
59             Edge e = edge[j];
60             if(!p[e.to]&&dist[e.to]>dist[k]+e.v)
61             {
62                 dist[e.to] = dist[k]+e.v;
63             }
64         }
65     }
66 }