最短路径之Dijkstra算法

第一行为源点个数，边的个数m

接下来m行为a->b和权值

6 9
1 2 1
1 3 12
2 3 9
2 4 3
3 5 5
4 3 4
4 5 13
4 6 15
5 6 4

求最短路径

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<queue>
 4 using namespace std;
 5 #define inf 1000000000
 6 int e[100][100], dis[100];
 7 int main()
 8 {
 9     int book[100] = { 0 };
10     int n, m;
11     scanf("%d %d", &n, &m);
12     for (int i = 1; i <= n; i++)
13         for (int j = 1; j <= n; j++)
14         {
15             if (i == j) e[i][j] = 0;
16             else e[i][j] = inf;
17         }
18     for (int i = 0; i < m; i++)
19     {
20         int a, b, c;
21         cin >> a >> b >> c;
22         e[a][b] = c;
23     }
24     for (int i = 1; i <= n; i++)
25         dis[i] = e[1][i];
26     book[1] = 1;
27     for (int i = 1; i < n; i++)
28     {
29         int temp = inf,mark = 0;
30         for (int j = 1; j <= n; j++)
31         {
32             if (book[j] == 0 && dis[j] < temp)
33             {
34                 temp = dis[j];
35                 mark = j;
36             }
37         }
38         book[mark] = 1;
39         for (int i = 1; i <= n; i++)
40         {
41             if (e[mark][i] < inf)
42             {
43                 dis[i] = min(e[mark][i] + dis[mark], dis[i]);
44             }
45         }
46     }
47     for (int i = 1; i <= n; i++)
48         cout << dis[i] << endl;
49 
50     return 0;
51 }