NK05DSRT - Sa mạc

Tác giả: RR

Ngôn ngữ: Pascal

//Wishing myself a happy lunar new year with a lot of accept solutions
//Written by Nguyen Thanh Trung (algorithm by taek)
{$R+,Q+}
uses math;
const
  FINP='';
  FOUT='';
  MAXN=101;
  oo=1000000000000;
type
  list=^node;
  node=record
         u,c:longint;
         next:list;
       end;
var
  f1,f2:text;
  test,n,gh,hsize:longint;
  ke:array[1..MAXN] of list;
  h,hpos:array[1..MAXN] of longint;
  need:array[1..MAXN] of int64;
procedure openF;
begin
  assign(f1,FINP); reset(f1);
  assign(f2,FOUT); rewrite(f2);
end;
procedure closeF;
begin
  close(f1); close(f2);
end;
procedure add(u,c:longint; var a:list);
var p:list;
begin
  new(p); p^.u:=u; p^.c:=c;
  p^.next:=a; a:=p;
end;
procedure inp;
var
  m,u,v,c:longint;
begin
  read(f1,n,m,gh);
  fillchar(ke,sizeof(ke),0);
  for m:=1 to m do
    begin
      read(f1,u,v,c);
      add(u,c,ke[v]);
      add(v,c,ke[u]);
    end;
end;
procedure ans;
begin
  if need[1]=oo then writeln(f2,-1)
  else writeln(f2,need[1]);
end;
//Xu ly heap
procedure swap(var a,b:longint);
var
  temp:longint;
begin
  temp:=a; a:=b; b:=temp;
end;
procedure downHeap(i:longint);
var
  j:longint;
begin
  j:=i<<1;
  while (j<=hsize) do
    begin
      if (j<hsize) and (need[h[j+1]]<need[h[j]]) then inc(j);
      if (need[h[j]]<need[h[i]]) then
        begin
          swap(hpos[h[i]],hpos[h[j]]);
          swap(h[i],h[j]);
        end;
      i:=j; j:=i<<1;
    end;
end;
procedure upHeap(i:longint);
var
  j:longint;
begin
  j:=i>>1;
  while (i>1) and (need[h[i]]<need[h[j]]) do
    begin
      swap(hpos[h[i]],hpos[h[j]]);
      swap(h[i],h[j]);
      i:=j; j:=i>>1;
    end;
end;
procedure push(u:longint);
begin
  inc(hsize); h[hsize]:=u; hpos[u]:=hsize;
  upHeap(hsize);
end;
procedure pop(var u:longint);
begin
  u:=h[1]; hpos[u]:=0;
  swap(h[1],h[hsize]); hpos[h[1]]:=1;
  dec(hsize); downHeap(1);
end;
procedure solve;
var
  u,v:longint;
  x,one,c:int64;
  p:list;
begin
  for u:=1 to n-1 do need[u]:=oo; need[n]:=0;
  hsize:=0; fillchar(hpos,sizeof(hpos),0); push(n);
  while hsize>0 do
    begin
      pop(u);
      p:=ke[u];
      while p<>nil do
        begin
          v:=p^.u; c:=p^.c; p:=p^.next;
          //Khong the qua canh (u,v)
          if c>gh then continue;
          //Chi can luong <=gh-c de den dich (chi can qua canh (u,v) 1 lan)
          if need[u]<=gh-c then x:=need[u]+c
          else if gh-c<<1<=0 then continue
          else
            begin
              //Luong nuoc chuyen duoc trong 1 lan
              one:=(need[u]-(gh-c)-1) div (gh-c<<1)+1;
              //Luong nuoc can
              x:=need[u]+(one<<1+1)*c;
            end;
          if x<need[v] then
            begin
              need[v]:=x;
              if hpos[v]=0 then push(v)
              else upHeap(hpos[v]);
            end;
        end;
    end;
end;
begin
  openF;
  read(f1,test);
  for test:=1 to test do
    begin
      inp;
      solve;
      ans;
    end;
  closeF;
end.

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