将给出的答案解释为两个数组

Question

这意味着什么

最小生成的边树在数组mst(大小为n-1乘2(中返回

？

当我运行程序时，在某个时刻显示

这两个阵列

然而，我不知道如何将其解释为最小生成树的边。

我该如何处理边缘？有没有办法画出这个答案？

有人能帮忙吗？

这就是代码。

function [mst, cost] = prim(A)
[n,n] = size(A);                           
A, n, pause,
if norm(A-A','fro') ~= 0 ,                 
disp(' Error:  Adjacency matrix must be symmetric ') 
return,
end;
intree = [1];  number_in_tree = 1;  
number_of_edges = 0;
notintree = [2:n]';  number_notin_tree= n-1;
in = intree(1:number_in_tree),                
out = notintree(1:number_notin_tree),
pause, 
while number_in_tree < n,
mincost = Inf;                             
for i=1:number_in_tree,               
for j=1:number_notin_tree,
ii = intree(i);  jj = 
notintree(j);
if A(ii,jj) < mincost, 
mincost = A(ii,jj); jsave = j; 
iisave = ii; jjsave = jj;   
end;
end;
end;
number_of_edges = number_of_edges +1;      
mst(number_of_edges,1) = iisave;            
mst(number_of_edges,2) = jjsave;
costs(number_of_edges,1) = mincost;
number_in_tree = number_in_tree + 1;        
intree = [intree; jjsave];                  
for j=jsave+1:number_notin_tree,            
notintree(j-1) = notintree(j);
end;
number_notin_tree = number_notin_tree - 1;  
in = intree(1:number_in_tree),              
out = notintree(1:number_notin_tree), 
pause,
end;
disp(' Edges in minimum spanning tree and their costs: ')
[mst  costs]                                 
cost = sum(costs)

Answer 1

边可以通过其连接的两个顶点唯一标识。mst的每一行包含到跨越边的两个顶点的两个索引。

输入图由一组顶点和连接它们的边组成，表示为邻接矩阵A。如果A(i,j)为真，则顶点i和j是相邻的(即共享边(。在输出矩阵mst中，该边缘将由mst(index,:) = [i,j]表示。