Dijkstra board

  . 1 #include <cstdio>
   2 #include <CString>
   . 3 #include <algorithm>
   . 4 #include <the iostream>
   . 5 #include <Vector>
   . 6 #include <Stack>
   . 7  the using  namespace STD;
   . 8  const  int MAXN = 505 ;
   . 9  const  int INF = 0x3f3f3f3f ;
 10  int n-, m, S, D;
 . 11  int SP [MAXN]; // number of relief 
12 is  int DJ [MAXN]; // distance shortest distance starting point 
13 is  intVIS [MAXN]; // the inclusion of a set of 
14  int Road [MAXN]; // shortest path number of 
15  int path [MAXN]; // prefix node 
16  int Total [MAXN]; // total number of relief 
. 17  struct City
 18 is  {
 . 19      int E;
 20 is      int len;
 21 is  };
 22 is Vector <City> VE [MAXN];
 23 is  // initialization 
24  void the init ()
 25  {
 26 is      Memset (VIS, 0 , the sizeof (VIS));
 27     memset (total,0,sizeof(total));
 28     memset (road,0,sizeof(road)); 
 29     memset (path,-1,sizeof(path));
 30     for (int i=0;i<n;i++)
 31         dj[i]=INF;
 32     dj[s]=0;
 33     total[s]=sp[s];
 34     road[s]=1;
 35 }
 36 void djst ()
 37 {
 38 is      the while ( . 1 )
 39      {
 40         int Maxx = INF;
 41 is         int U = - . 1 ;
 42 is         for ( int I = 0 ; I <n-; I ++ )
 43 is              IF (DJ [I] <Maxx &&! VIS [I])
 44 is              {
 45                 Maxx = DJ [I];
 46 is                 u = I;
 47              }
 48         IF (u == - . 1 ) // if u still or equal to -1 DESCRIPTION all the points have been minimized, the loop end 
49           BREAK ;
 50         VIS [u] = . 1 ; // otherwise an intermediary to a point u as to extend to the periphery 
51 is         for ( int I = 0 ; I <VE [u] .size (); I ++ )
 52 is         {
 53 is             int V = VE [u] [I] .E; // to point u through point u to find edges point v, v point to diffuse 
54 is             IF (VIS [v]!) // If v has not visited point 
55             {
 56 is                 IF (DJ [v]> DJ [U] + VE [U] [I] .LEN) // if the direct v to v farther than the point reached by u, is updated at 
57 is                 {
 58                    DJ [v] DJ = [U ] + VE [U] [I] .LEN;
 59                   path [v] = u; // record what path, the precursor is v u 
60                    Road [v] = Road [u]; // to v shortest path and the shortest path number to u as strip 
61 is                    Total [ v] = Total [u] + SP [v]; // to rescue the number v is the number of the number to u + v relief 
62 is   
63 is                 }
 64                 the else  IF (DJ [v] == DJ [u] + VE [ u] [I] .LEN) // if directly to V and by the same length from u to v, 
65                 {
 66                     // here to be written out if statement 
67                      Road [v] = + Road [u]; // that the v is the shortest path will add the number to the number of shortest path u strip 
68                      IF (Total [u] + SP [v]> Total [v]) // update recovery and path number, because the number of rescue is the second title keyword
69                      {
 70                         path [v] = U; // If updated, then it must be a precursor v U 
71 is                         Total [v] = Total [U] + SP [v];
 72                      }
 73 is                 }
 74             }
 75         }
 76      }
 77  }
 78  // output function 
79  void output ()
 80  {
 81      the printf ( " % D% D \ n- " , Road [D], Total [D]);
 82      int TEMP = D;
 83      Stack <int>ss;
 84     while (path[temp]!=-1)
 85     {
 86         ss.push(temp);
 87         temp=path[temp];
 88     }
 89     ss.push(s);
 90     while (!ss.empty())
 91     {
 92         printf("%d%c",ss.top(),ss.size()==1? '\n':' ');
 93         ss.pop();
 94     }
 95 }
 96 int main()
 97 {
 98     scanf("%d%d%d%d",&n,&m,&s,&d);
 99     for (int i=0;i<n;i++)
100         scanf("%d",&sp[i]);
101     init();
102     for (int i=0;i<m;i++)
103     {
104         int x,y,len;
105         scanf("%d%d%d",&x,&y,&only);
106          City temp1, temp2 of;
 107          temp1.e = Y; temp1.len = len;
 108          temp2.e = X; temp2.len = len;
 109          VE [X] .push_back (temp1); // undirected graph 
110          VE [Y] .push_back (temp2 of);
 111      }
 112      DJST ();
 113      Output ();
 114      return  0 ;
 115 }