UVa 10986:sending Email (dijkstra optimization, SPFA)

Link:

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=1927

Topic:

Problem E
Sending email
Time Limit:3 seconds

"A new internet watchdog is creating a stir in
Springfield. Mr. X, if, is he real name, has
Come up with a sensational scoop. "
Kent Brockman

There are n SMTP servers connected by network cables. Each of the M cables connects two computers and has a certain latency-measured in milliseconds required to send an email m Essage. What is the shortest time required to send a message from server S to server T along a sequence of cables? Assume that there are no delay incurred at any of the servers.

Input
The "a" of input gives the number of cases, N. n test Cases follow. Each one is starts with a line containing n (2<=n<20000), M (0<=m<50000), S (0<=s<n) and T (0<=t<n). S!=t. The next m lines'll each contain 3 integers:2 different servers (in the range [0, n-1]) this are connected by a Bidirec tional Cable and the latency, W, along this cable (0<=w<=10000).

Output
For each test case, output the ' line ' case #x: ' followed by the number of milliseconds required to send a message from S T. Print "Unreachable" if there is no route the from S to T.
Sample Input

3
2 1 0 1
0 1 100
3 3 2 0
0 1 100
0 2 200
1 2 50
2 0 0 1

Sample Output
Case #1:100
Case #2:150
Case #3: Unreachable

Problemsetter:igor Naverniouk

The main effect of the topic:

For a graph, find the minimum distance from the s point to the T point.

Analysis and Summary:

Naked the shortest, but n too large is obviously not used adjacency matrix, need to use adjacency table + Priority queue optimization.

Code:

1. Dijkstra, 0.148s

#include <cstdio> #include <cstring> #include <utility> #include <queue> using namespace std;  
    
typedef pair<int,int>pii;  
    
Priority_queue<pii,vector<pii>,greater<pii> >q;  
const int N = 100005;  
    
const int INF = 1000000000;  
int n, m, Beg, end, K;  
int head[n], next[n], u[n], v[n], w[n], d[n];  
     
BOOL Vis[n];  
    inline void Read_graph () {scanf ("%d%d%d%d", &n,&m,&beg,&end);  
    Memset (Head,-1, sizeof (head));  
        for (int e=1; e<=m; ++e) {scanf ("%d%d%d", &u[e],&v[e],&w[e]);  
        U[e+m]=v[e], v[e+m]=u[e], w[e+m]=w[e];  
        Next[e] = head[u[e]];  
        Head[u[e]] = e;  
        NEXT[E+M] = head[u[e+m]];  
    Head[u[e+m]] = e+m;  
    } inline void Dijkstra (int src) {memset (Vis, 0, sizeof (VIS));  
    for (int i=0; i<n; ++i) d[i] = INF;  
    D[SRC] = 0;  
    Q.push (Make_pair (d[src], SRC));  
        while (!q.empty ()) {PII u = q.top ();  
        Q.pop ();  
        int x = U.second;  
        if (vis[x]) continue;  
        Vis[x] = true;  
            for (int e=head[x]; e!=-1 e=next[e]) if (D[v[e]] > D[x]+w[e]) {d[v[e]] = d[x]+w[e];  
        Q.push (Make_pair (D[v[e]], v[e));  
    int main () {int t,cas=1;  
    scanf ("%d", &t);  
        while (t--) {read_graph ();  
        Dijkstra (Beg);  
        printf ("Case #%d:", cas++);  
        if (d[end]!=inf) printf ("%d\n", D[end]);  
    Else puts ("unreachable");  
return 0; }