toj 2819 Travel

2819.   Travel

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Time Limit: 1.0 Seconds    Memory Limit: 65536K
Total Runs: 1042    Accepted Runs: 458

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There are N cities in this country, and some of them are connected by roads. Given two cities, can you find the shortest distance between them?

Input

The first line contains the number of test cases, then some test cases followed.

The first line of each test case contain two integers N and M (3 ≤ N ≤ 1000, M ≥ 0), indicating the number of cities and the number of roads. The next line contain two numbers S and T (S ≠ T), indicating the start point and the destination. Each line of the following M lines contain three numbers A_i, B_i and C_i (1 ≤ A_i,B_i ≤ N, 1 ≤ C_i ≤ 1000), indicating that there is a road between city A_i and city B_i, whose length is C_i. Please note all the cities are numbered from 1 to N.

The roads are undirected, that is, if there is a road between A and B, you can travel from A to B and also from B to A.

Output

Output one line for each test case, indicating the shortest distance between S and T. If there is no path between them, output -1.

Sample Input

2
3 3
1 3
1 2 10
2 3 20
1 3 50
3 1
1 3
1 2 10

Sample Output

30
-1

Problem Setter: RoBa

Source: TJU Programming Contest 2007 Submit   List    Runs   Forum   Statistics

---

#include  < iostream >
#include  < queue >
#define  MAX 1002
using   namespace  std;
typedef  struct  node
{
     int  di;
     int  heap;
    node(){}
    node( int  h, int  d)
    {
        heap = h;
        di = d;
    }
    friend  bool   operator   < (node a,node b)
    {
         return  a.di > b.di;
    }
}Point;
priority_queue < Point > Q;
Point temp;
int  n,m,dis[MAX],t,s,e;
int  edges[MAX][MAX];
bool  mark[MAX];
void  Init()
{
     int  i,j,a,b,ss;
    scanf( " %d%d " , & n, & m);
    scanf( " %d%d " , & s, & e);
     for (i = 1 ;i <= n;i ++ )
         for (j = 1 ;j <= n;j ++ )
            edges[i][j] =- 1 ;
     while (m -- )
    {
        scanf( " %d%d%d " , & a, & b, & ss);
         if (edges[a][b] ==- 1 || s < edges[a][b])
        {
            edges[a][b] = ss;
            edges[b][a] = ss;
        }
    }
}
void  Dijkstra( int  s)
{
     int  i,j,k;
     while ( ! Q.empty())
        Q.pop();
     for (i = 1 ;i <= n;i ++ )
    {
        dis[i] = edges[s][i];
         if (dis[i] !=- 1 )
        {
            Q.push(node(i,dis[i]));
        }
        mark[i] = false ;
    }
    mark[s] = true ;
     for (i = 1 ;i < n;i ++ )
    {
        k =- 1 ;
         while ( ! Q.empty())
        {
            temp = Q.top();
            Q.pop();
             if ( ! mark[temp.heap])
            {
                k = temp.heap;
                 break ;
            }
        }
         if (k ==- 1 )
        {
            printf( " -1\n " );
             return  ;
        }
         for (j = 1 ;j <= n;j ++ )
        {
             if (mark[j] || edges[k][j] ==- 1 )
                 continue ;
             if (dis[k] + edges[k][j] < dis[j] || dis[j] ==- 1 )
            {
                dis[j] = dis[k] + edges[k][j];
                Q.push(node(j,dis[j]));
            }
        }
    }
    printf( " %d\n " ,dis[e]);
}
int  main()
{
    cin >> t;
     while (t -- )
    {
        Init();
        Dijkstra(s);
    }
     return   0 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/14/1456968.html