Just as a reminder with simple explanation.

Path Reconstruction #

Use vector<int> pre[N] to record the previous vertices of all the vertices in the shortest path(s). When updating the distance to vertex $v$, if the current distance is better, discard the previous record and let the current vertex be the previous vertex of $v$. If the distance is the same, just add the current vertex to pre[v].

for(pii it:E[u]){
    ll v=it.S,cost=it.F;
    if(!vis[v]&&dis[v]>dis[u]+cost){
        dis[v]=dis[u]+cost;
        pre[v].clear();
        pre[v].pb({cost,u});
        q.push({dis[v],v});
    }else if(dis[v]==dis[u]+cost)
        pre[v].pb({cost,u});
}

Number of shortest paths #

Similar to recording the path, if the distance is better then let the number be one. If the same, plus 1.

if(!vis[v]&&dis[u]+cost<dis[v]){
    cnt[v]=1;
    dis[v]=dis[u]+cost;
}else if(dis[u]+cost==dis[v]){
    cnt[v]++;
}