Dijkstra最小步长最短路径

我需要实现一个程序，给定一个有向图，在弧线处具有正成本，打印从x到y的最小成本以及从x到y的所有路径的最小步数。我写了一个程序来做这个，但有时它选择的路径不是最小步数。下面是我的代码:

#include <iostream>
#include <vector>
#include <queue>
#include <climits>
using namespace std;
typedef pair<int, int> Warc;
typedef vector<vector<Warc>> Wgraph;
int dijkstra(const Wgraph& g, int x, int y, vector<int>& p) {
  int n = g.size();
  vector<int> d(n, INT_MAX); //costs
  p = vector<int>(n, -1); //parent of each vertice
  d[x] =0;
  vector<bool> S(n, false);
  priority_queue<Warc, vector<Warc>, greater<Warc>> Q;
  Q.push(Warc(0, x));
  while(not Q.empty()) {
    int u = Q.top().second;
    Q.pop();
    if(u == y) return d[u];
    if(not S[u]) {
      S[u] = true;
      for(Warc a : g[u]) {
        int v = a.second;
        int c = a.first;
        if(d[v] > d[u] + c) {
         d[v] = d[u] + c;
         p[v] = u;
         Q.push(Warc(d[v], v));
       }
      }
    }
  }
  return -1;
}
int main() {
  int n; //# of vertices
  while(cin >> n) {
    int m; //# of arcs
    cin >> m;
    Wgraph g(n);
    for(int i = 0; i<m; ++i) {
      int u, v;
      int c;
      cin >> u >> v >> c; // arc: u->v with cost C
      g[u].push_back(make_pair(c, v));
    }
    int x, y;
    cin >> x >> y; //origin and end
    vector<int> p; //p[i]= parent of vertice i
    int path = dijkstra(g, x, y, p);
    if(path == -1) cout << "no path from " << x << " to " << y << endl;
    else {
        int s = 0;
        int i = y;
        while(i != x) {
            ++s;
            i = p[i];
        }
        cout << "cost " << path-s << ", " << s << " step(s)" << endl;
    }
  }
}

谢谢。