pascal编写的floyd算法求两点间的最短路径

我的程序：

program floyd;
var
k,n,i,j,x,max:integer;
a:array[1..5,1..5]of integer;
path:array[1..5,1..5]of integer;
procedure print(i,j:integer);    输出vi到vj最短路径的顶点序列
var k:integer;
begin
   k:=path[i,j];
   if k<>0 then
      begin
        print(i,k);
        write(k);
        print(k,j);
      end;
end;
begin
assign(input,'floyd.in');
reset(input);
assign(output,'floyd.out');
rewrite(output);
readln(n);
max:=maxint;
for i:=1 to n do    //初始化邻接矩阵
 begin
  for j:=1 to n do
    begin
     read(x);
     if x<>0 then a[i,j]:=x else
       if (x=0) and (i=j) then a[i,j]:=x else a[i,j]:=maxint;
    end;
  readln;
 end;
 for i:=1 to n do      //初始化path
    for j:=1 to n do
       if (a[i,j]<>0) and (a[i,j]<>maxint) then path[i,j]:=i;

for k:=1 to n do    枚举加入图中的每个点
  for i:=1 to n do
    for j:=1 to n do
      if (a[i,j]<>0) and (j<>k) and (i<>k) and (a[i,k]<>max) and (a[k,j]<>max)  then
       if (a[i,j]>a[i,k]+a[k,j]) then
                      begin
                      a[i,j]:=a[i,k]+a[k,j];      更新顶点i到j的最短路
                      path[i,j]:=k;
                      end;

for i:=1 to n do
  begin
    for j:=1 to n do
    if a[i,j]=max then write(0:4) else write(a[i,j]:4);
    writeln;
  end;
print(2,5);             递归求出v2到v5最短路的顶点序列
close(input);
close(output);
end.

问题：我把子过程去掉的话，程序运行输出a[i,j]的最短路径结果是正确的，但我把输出的子程序加进去之后，编译的时候会栈溢出 stack overflow error 请问是怎么一回事呢？请高手不吝赐教~~~~

//我的程序：

program floyd;
var
k,n,i,j,x,max:integer;
a:array[1..5,1..5]of integer;
path:array[1..5,1..5]of integer;

procedure print(i,j:integer);    //输出vi到vj最短路径的顶点序列
var k:integer;
begin
 k:=path[i,j];
 if k=0 then exit;
 print(i,k);
 write(' ',k);
 print(k,j);
end;

begin
assign(input,'floyd.in');reset(input);
assign(output,'floyd.out');rewrite(output);
readln(n);
max:=maxint;
for i:=1 to n do    //初始化邻接矩阵
 begin
  for j:=1 to n do
    begin
    read(x);
    if x<>0 then a[i,j]:=x else
    if (x=0) and (i=j) then a[i,j]:=x else a[i,j]:=maxint;
    end;
  readln;
 end;

{ for i:=1 to n do    //初始化path
    for j:=1 to n do
    if (a[i,j]<>0) and (a[i,j]<>maxint) then path[i,j]:=i; }   把初始化删掉，不然会死循环。

for k:=1 to n do    //枚举加入图中的每个点
  for i:=1 to n do
    for j:=1 to n do
    if (a[i,j]<>0) and (j<>k) and (i<>k) and (a[i,k]<>max) and (a[k,j]<>max)  then
    if (a[i,j]>a[i,k]+a[k,j]) then
    begin
    a[i,j]:=a[i,k]+a[k,j];    // 更新顶点i到j的最短路
    path[i,j]:=k;
    end;

for i:=1 to n do
  begin
    for j:=1 to n do
    if a[i,j]=max then write(0:4) else write(a[i,j]:4);
    writeln;
  end;

write(2);print(2,5);writeln(' ',5);    //递归求出v2到v5最短路的顶点序列

    
close(input);
close(output);
end.

给个测试数据，这样好检查