Minimum Spanning Tree

For a given weighted graph G=(V,E), find the minimum spanning tree (MST) of G

and print total weight of edges belong to the MST.

Input

In the first line, an integer n

denoting the number of vertices in G is given. In the following n lines, a n×n adjacency matrix A which represents G is given. aij denotes the weight of edge connecting vertex i and vertex j. If there is no edge between i and j, aij

is given by -1.

Output

Print the total weight of the minimum spanning tree of G

.

Constraints

  • 1≤n≤100

(if aij≠−1

  • is a connected graph

Sample Input 1

5
 -1 2 3 1 -1
 2 -1 -1 4 -1
 3 -1 -1 1 1
 1 4 1 -1 3
 -1 -1 1 3 -1

Sample Output 1

5

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

 1 #include<iostream>
 2 using namespace std;
 3 static const int MAX=100;
 4 static const int INFTY=(1<<21);
 5 static const int WHITE =0;
 6 static const int GRAY=1;
 7 static const int BLACK=2;
 8 int n,M[MAX][MAX];
 9 int prim()
10 {
11     int u,minv;
12     int d[MAX],p[MAX],color[MAX];
13     for(int i=0;i<n;i++)
14     {
15         d[i]=INFTY;
16         p[i]=-1;
17         color[i]=WHITE;
18     }
19 
20     d[0]=0;
21     while(1)
22     {
23         minv=INFTY;
24         u=-1;
25         for(int i=0;i<n;i++)
26         {
27             if(minv>d[i]&&color[i]!=BLACK)
28             {
29                 u=i;
30                 minv=d[i];
31             }
32         }
33         if(u==-1)
34         {
35             break;
36         }
37         color[u]=BLACK;
38         for(int v=0;v<n;v++)
39         {
40             if(color[v]!=BLACK&&M[u][v]!=INFTY)
41             {
42                 if(d[v]>M[u][v])
43                 {
44                     d[v]=M[u][v];
45                     p[v]=u;
46                     color[v]=GRAY;
47                 }
48             }
49         }
50     }
51     int sum=0;
52     for(int i=0;i<n;i++)
53     {
54         if(p[i]!=-1)
55             sum+=M[i][p[i]];
56 
57     }
58     return sum;
59 }
60 int main()
61 {
62     cin>>n;
63     for(int i=0;i<n;i++)
64     {
65         for(int j=0;j<n;j++)
66         {
67             int e;
68             cin>>e;
69             M[i][j]=(e==-1)?INFTY:e;
70         }
71     }
72     cout<<prim()<<endl;
73     return 0;
74 }

 

G

posted @ 2020-12-02 19:53  BlackSnow  阅读(129)  评论(0)    收藏  举报