Ou Tao shortest recording the shortest path []

Topic links: https://ac.nowcoder.com/acm/contest/1168/C

Is the most common short-circuit, built from the point of view of even less than the two-way side m, since that write a lot but do not know how the shortest path shortest of record. Record enough in relaxation, the relaxation is recorded each time point was slack precursor, so that when a point repeatedly relaxation, you can update several times its predecessor, can not slack when it is the point on the shortest path, that is, the last recorded slack precursor.

And from the end traversing precursor, present in the array to output a reverse shortest path through the output point.

code show as below:

  1 #include<bits/stdc++.h>
  2 #define mem(a,b) memset(a,b,sizeof(a))
  3 const int MAXN = 650;
  4 const double inf = 0x3f3f3f3f * 1.0;
  5 using namespace std;
  6 
  7 int n, k;
  8 double m, dis[MAXN];
  9 int head[MAXN], cnt, vis[MAXN];
 10 int pre[MAXN]; 
 11 
 12 struct Node
 13 {
 14     double x, y, z;
 15 }node[MAXN];
 16 
 17 struct Edge
 18 {
 19     int to, next;
 20     double w;
 21 }edge[MAXN * MAXN];
 22 
 23 struct N
 24 {
 25     int pot;
 26     double dis;
 27     bool operator < (const N &a)const
 28     {
 29         return dis > a.dis;
 30     }
 31 }no;
 32 
 33 void add(int a, int b, double c)
 34 {
 35     cnt ++;
 36     edge[cnt].to = b;
 37     edge[cnt].next = head[a];
 38     edge[cnt].w = c;
 39     head[a] = cnt;
 40 }
 41 
 42 void dij()
 43 {
 44     priority_queue<N> Q;
 45     no.pot = 0, no.dis = 0;
 46     fill(dis, dis + n + 2, inf), mem(vis, 0);
 47     dis[0] = 0;
 48     Q.push(no);
 49     N a;
 50     while(!Q.empty())
 51     {
 52         a = Q.top();
 53         Q.pop();
 54         if(vis[a.pot])
 55             continue;
 56         vis[a.pot] = 1;
 57         for(int i = head[a.pot]; i != -1; i = edge[i].next)
58          {
 59              
60              int to = Edge [I] .to;
 61 is              IF (DIS [to]> DIS [a.pot] + Edge [I] .W)
 62 is              {
 63 is                  DIS [to] DIS = [a.pot] + Edge [I] .W;
 64                  no.pot = to, no.dis = DIS [to];
 65                  pre [to] = a.pot;   // relaxation record is updated, it updates the last recorded some slack to the minimum point on the path 
66                  Q.push (NO);
 67              }
 68          }
 69      }
 70  }
 71 is  
72  int main ()
 73 {
 74     mem(head, -1), cnt = 0;
 75     scanf("%d%lf", &n, &m);
 76     scanf("%lf%lf%lf%lf%lf%lf", &node[0].x, &node[0].y, &node[0].z, &node[n + 1].x, &node[n + 1].y, &node[n + 1].z); //起点 终点 
 77     for(int i = 1; i <= n; i ++)
 78         scanf("%lf%lf%lf", &node[i].x, &node[i].y, &node[i].z);
 79     for(int i = 0; i < n + 1; i ++)
 80     {
 81         for(int j = i + 1; j <= n + 1; j ++)
 82         {
 83             double jl = (node[i].x - node[j].x) * (node[i].x - node[j].x) + (node[i].y - node[j].y) * (node[i].y - node[j].y) + (node[i].z - node[j].z) * (node[i].z - node[j].z);
 84             if(jl <= m * m)
 85             {
 86                 add(i, j, sqrt(jl));
 87                 add(j, i, sqrt(jl));
 88             }
 89         }
 90     }
 91     dij();
 92     if(dis[n + 1] != inf * 1.0)
 93     {
 94         printf("%.3lf\n", dis[n + 1]);
 95         printf("Start ");
 96         int p, len = 0, ans[MAXN];
 97         p = n + 1; // end 
98          the while (P =! 0 ) // when the starting point is not equal to 
99          {
 100              ANS [len ++] = P;
 101              P = pre [P];
 102          }
 103          for ( int I = len . 1 ; I> 0 ; i-- )
 104              the printf ( " % D " , ANS [I]);
 105          the printf ( " End \ n- " );
 106      }
 107      the else 
108         printf("-1\n");
109     return 0;
110 }

Recording the shortest path passes through a point