HDU 1535 Invitation Cards (Reverse composition +dijkstra Priority queue optimization)

http://acm.hdu.edu.cn/showproblem.php?pid=1535

Main topic: The shortest path from 1 to any point and the shortest point to the starting point 1

Problem-solving ideas: a positive composition to find the shortest way, and then the direction of the map to change, reverse composition and then find the shortest way, the last two times accumulated and on the line.

Considering the amount of data to the point, Dijkstra need to use priority queue optimization.

First recall Dijkstra common practice is:

Clear all points of the label
set d[0]=0, other D[i]=inf
loop n times {
    in all unlabeled nodes, select the node with the lowest D value x to the
    node x mark
    for all edges starting from X (x, y), update d[y]=min{d [Y],d[x]+w (x, y)}
}    

Dijkstra is primarily optimized for the insertion (update) of numeric values and the removal of a minimum of two operations, so take into account the heap.

Dijkstra Priority Queue Optimization idea: In the STL with priority_queue< node > Q instead of the heap, each vertex current minimum distance with the heap to maintain, in the shortest distance to update, the corresponding element to the direction of the root to meet the nature of the heap. The minimum value that is removed from the heap each time is the next vertex to be used, so that the elements in the heap have O (V), and the update and fetch operations have O (E), so the complexity of the algorithm is O (ELOGV)

The specific code is implemented as follows:

#include <iostream> #include <cstring> #include <queue> #define INF 0x3f3f3f3f using namespace std;
const int maxn=1000005;
    struct node{int v,len;
    BOOL operator < (const node&a) const{return len>a.len; 
}
};
int N,M,DIS[MAXN],VIS[MAXN];
vector<node>e[maxn]; 
vector<node>e1[maxn];
    void Dijkstra (vector<node> g[])//dijkstra priority queue Optimization {memset (vis,0,sizeof (VIS));
    for (int i=0;i<=n;i++) Dis[i]=inf;
    dis[1]=0;
    priority_queue<node>q;
    Q.push (node{1,0}); while (!
        Q.empty ()) {node now=q.top ();
        Q.pop ();
        int v=now.v;
        if (Vis[v]) continue;
        Vis[v]=1;
            for (int i=0;i<g[v].size (); i++) {int u=g[v][i].v,len=g[v][i].len;
                if (!vis[u]&&dis[u]>dis[v]+len) {Dis[u]=dis[v]+len;
            Q.push (Node{u,dis[u]}); }}}} int main () {int t,a,b,C
    cin>>t;
        while (t--) {cin>>n>>m;
            for (int i=0;i<=n;i++) {e[i].clear ();
        E1[i].clear ();
            } while (m--) {cin>>a>>b>>c;   E[a].push_back (Node{b,c});    Forward graph E1[b].push_back (node{a,c});
        Reverse build side} int ans=0;
        Dijkstra (e);
        for (int i=1;i<=n;i++) ans+=dis[i];
        Dijkstra (E1);
        for (int i=1;i<=n;i++) ans+=dis[i];
    cout<<ans<<endl;
} return 0;  }