[Explanations] PAT 1003 Emergency

The main idea: Dijkstra + DP

If done most shorted count, this question is equivalent to a double experience.

That is, when running the Dijkstra, excluding a \ (t_i \) as the first \ (I \) the minimum number of short bar when the points, a \ (rs_i \) as the first \ (I \) points shortest point the maximum value and the right

Note the number of maintenance to be classified is not \ (dis \) update, the updated value before overwriting with the new value.

Also, the point number \ (0 \) ~ \ (. 1-n-\) , edges are bidirectional edge, make no mistake

Part of the code:

struct wtf{ 
    int t,r,rs;
}r[505];

struct node{
    int u,d;
    friend bool operator<(const node &a,const node &b){
        return a.d>b.d;
    }
};

void Dijkstra(){
    for(int i=0;i<=n-1;i++) dis[i]=inf;
    dis[c1]=0, r[c1].t = 1;
    priority_queue<node>q;
    node x;
    x.u=c1,x.d=0;
    q.push(x);
    while(!q.empty()) {
        node fr = q.top();
        q.pop();
        int u = fr.u, d = fr.d;
        if(d != dis[u]) continue;
        for(int i = head[u]; i != -1; i = e[i].next) {
            int v = e[i].to, w = e[i].w;
            if(dis[u] + w <= dis[v]) {
                bool flag = 0;
                if(dis[u] + w < dis[v])
                    r[v].t = r[u].t,
                    flag = 1;
                else if(dis[u] + w == dis[v])
                    r[v].t = r[v].t + r[u].t;
                    
                r[v].rs = max(r[u].rs + r[v].r, r[v].rs);
                dis[v] = dis[u] + w;
                if(flag) {
                    node y;
                    y.u = v, y.d = dis[v];
                    q.push(y);
                }
            }
        }
    }
}

PS: to help people see when they change the code title, the third year in the first knock Code ,,, Middle School continue to increase (tui) oil (fei)!