用vb.netl编写的floyd算法求两点间的最短路径,怎么输出path经过的顶点序列?

用vb.netl编写的floyd算法求两点间的最短路径,怎么输出path经过的顶点序列?

Function Min(x() as integer,y() as integer) as doubledim i,j,k,a dim m() as doubledim s() as stringdim mins as stringredim m(ubound(x),ubound(x))redim s(ubound(x),ubound(x))for i=1 to ubound(x)-1 '从起始点0点到i点的距离m(i,0)=((x(i)-x(0))^2+(y(i)-y(0))^2)^0.5s(i,0)=0- & cstr(i)next'从起始点开始经过K个点后到达i点的最短距离m(i,k),s为各点的连线如0-3-2-1-4for k=1 to ubound(x)-2for i=1 to ubound(x)-1m(i,k)=10^307for j=1 to ubound(x)-1if instr(s(j,k-1),cstr(i))=0 then'避免重复走一点a=((x(i)-x(j))^2+(y(i)-y(j))^2)^0.5if a+m(j,k-1)<m(i,k) thenm(i,k)=a+m(j,k-1)s(i,k)=s(j,k-1) & - & cstr(i)endifend ifnext nextnext'计算经过各点后到达最后一个点的最短距离min=10^307for j=1 to ubound(x)-1a=((x(ubound(x))-x(j))^2+(y(ubound(x))-y(j))^2)^0.5if a+m(j,ubound(x)-2)<min thenmin=a+m(j,ubound(x)-2)mins=s(j,ubound(x)-2) & - & cstr(ubound(x))end ifnextmsgbox 最短距离: & min & vbcrlf & 最短路径: & minsEnd functionprivate sub Command1_Clickdim x(5) as integerdim y(5) as integerdim m as doublex(0)=0y(0)=0x(1)=40y(1)=600.x(5)=1000y(5)=1000m=min(x,y)End sub

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