pku 1258 AgriNet

求一个图的最小生成树，用了prim,0ms，但前面WA了几次，这里进一步理解了prim，外部循环i只是控制次数，内部不用看i与j的值的关系，把d赋值成0，表示在树里，节省了一个布尔数组。

program pku1258(input,output);
var
   f      : array[0..200,0..200] of longint;
   d      : array[0..200] of longint;
   n      : longint;
   answer : longint;
procedure init;
var
   i,j : longint;
begin
   readln(n);
   fillchar(f,sizeof(f),0);
   for i:=1 to n do
      for j:=1 to n do
     read(f[i,j]);
   readln;
   for i:=1 to n do
      d[i]:=f[1,i];
end; { init }
procedure prim;
var
   i,j        : longint;
   min,minn : longint;
begin
   answer:=0;
   for i:=1 to n-1 do
   begin
      min:=maxlongint;
      for j:=1 to n do
     if (d[j]>0)and(d[j]<min) then
     begin
        min:=d[j];
        minn:=j;
     end;
      inc(answer,d[minn]);
      d[minn]:=0;
      for j:=1 to n do
     if (d[j]>0)and(f[minn,j]<d[j]) then
        d[j]:=f[minn,j];
   end;
   writeln(answer);
end; { prim }
begin
   while not eof do
   begin
      init;
      prim;
   end;
end.