HYSBZ 1050(队列-大小边比值最大的路径）

已知边，判断2点连通性

要用并查集……千万别搜啊~

Program ee;var   edge:array[1..10000,1..3] of longint;   s,t,n,m,i,j,pmax,pmin:longint;   father:array[1..1000] of longint;procedure swap(var a,b:longint);var   p:longint;begin   p:=a;   a:=b;   b:=p;end;procedure qsort(l,r:longint);var   i,j,m,p:longint;begin   i:=l;   j:=r;   m:=edge[(l+r) shr 1,3];   repeat      while edge[i,3]<m do inc(i);      while edge[j,3]>m do dec(j);      if i<=j then      begin         swap(edge[i,1],edge[j,1]);         swap(edge[i,2],edge[j,2]);         swap(edge[i,3],edge[j,3]);         inc(i);dec(j);      end;   until i>j;   if l<j then qsort(l,j);   if i<r then qsort(i,r);end;function gcd(a,b:longint):longint;begin   if b=0 then exit(a) else exit(gcd(b,a mod b));end;function getfather(x:longint):longint;begin   if father[x]=x then exit(x) else father[x]:=getfather(father[x]);   exit(father[x]);end;begin   pmax:=50000000;pmin:=1;   read(n,m);   for i:=1 to m do read(edge[i,1],edge[i,2],edge[i,3]);   read(s,t);   qsort(1,m);   i:=1;   for i:=1 to m do   begin      for j:=1 to n do father[j]:=j;      j:=i;      while (getfather(s)<>getfather(t)) and (j<=m) do      begin         if getfather(edge[j,2])<>getfather(edge[j,1]) then father[father[edge[j,2]]]:=father[father[edge[j,1]]];         inc(j);      end;      if getfather(s)=getfather(t) then      begin         dec(j);         if (pmax/pmin>edge[j,3]/edge[i,3]) then         begin            pmax:=edge[j,3];            pmin:=edge[i,3];         end;      end;   end;   if pmax=50000000 then writeln('IMPOSSIBLE')   else if (pmax mod pmin=0) then writeln(pmax div pmin)   else   begin      i:=gcd(pmax,pmin);      pmax:=pmax div i;pmin:=pmin div i;      writeln(pmax,'/',pmin);   end;end.