【NOIP2016练习&BZOJ2125】T3 sp （树上倍增，最短路）

题意：给一个N个点M条边的连通无向图，满足每条边最多属于一个环，有Q组询问，每次询问两点之间的最短路径。

对于80%的数据环的个数<=1

对于100%的数据N<=10000，Q<=10000

思路：bzoj原题

        只会80，正解听说是仙人掌

        对于80每个询问的最优解必定是一棵树或一棵树+一条边，因为环的个数<=1

        倍增LCA求距离，枚举树上加哪条边即可

        

var f,g:array[1..100000,0..20]of longint;
    head,vet,next,fa,dep,a,b,c,len:array[1..200000]of longint;
    n,i,tx,ty,tz,ans,w,tot,m,q,x,y,j:longint;

function min(x,y:longint):longint;
begin
 if x<y then exit(x);
 exit(y);
end;

procedure add(a,b,c:longint);
begin
 inc(tot);
 next[tot]:=head[a];
 vet[tot]:=b;
 len[tot]:=c;
 head[a]:=tot;
end;

procedure swap(var x,y:longint);
var t:longint;
begin
 t:=x; x:=y; y:=t;
end;

function lca(x,y:longint):longint;
var i,d:longint;
begin
 if dep[x]<dep[y] then swap(x,y);
 d:=dep[x]-dep[y];
 for i:=0 to 14 do
  if d and (1<<i)>0 then x:=f[x,i];
 for i:=14 downto 0 do
  if f[x,i]<>f[y,i] then
  begin
   x:=f[x,i]; y:=f[y,i];
  end;
 if x=y then exit(x);
 exit(f[x,0]);
end;

function clac(x,y:longint):longint;
var i,d:longint;
begin
 if dep[x]<dep[y] then swap(x,y);
 d:=dep[x]-dep[y];
 clac:=0;
 for i:=0 to 14 do
  if d and (1<<i)>0 then
  begin
   clac:=clac+g[x,i];
   x:=f[x,i];
  end;
end;

function ask(x,y:longint):longint;
var t:longint;
begin
 t:=lca(x,y);
 exit(clac(x,t)+clac(y,t));
end;

function find(k:longint):longint;
begin
 if fa[k]<>k then fa[k]:=find(fa[k]);
 exit(fa[k]);
end;

procedure dfs(u,pre:longint);
var i,e,v:longint;
begin
 for i:=1 to 14 do
 begin
  if dep[u]<1<<i then break;
  f[u,i]:=f[f[u,i-1],i-1];
  g[u,i]:=g[u,i-1]+g[f[u,i-1],i-1];
 end;
 e:=head[u];
 while e<>0 do
 begin
  v:=vet[e];
  if v<>pre then
  begin
   dep[v]:=dep[u]+1;
   f[v,0]:=u; g[v,0]:=len[e];
   dfs(v,u);
  end;
  e:=next[e];
 end;
end;

begin
 assign(input,'sp.in'); reset(input);
 assign(output,'sp.out'); rewrite(output);
 readln(n,m,q);
 for i:=1 to n do fa[i]:=i;
 for i:=1 to m do
 begin
  readln(tx,ty,tz);
  x:=find(tx); y:=find(ty);
  if x<>y then
  begin
   fa[y]:=x;
   add(tx,ty,tz);
   add(ty,tx,tz);
  end
   else begin inc(w); a[w]:=tx; b[w]:=ty; c[w]:=tz; end;
 end;
 dfs(1,-1);
 for i:=1 to q do
 begin
  read(tx,ty);
  ans:=ask(tx,ty);
  for j:=1 to w do
  begin
   ans:=min(ans,ask(tx,a[j])+c[j]+ask(ty,b[j]));
   ans:=min(ans,ask(ty,a[j])+c[j]+ask(tx,b[j]));
  end;
  writeln(ans);
 end;
 close(input);
 close(output);
end.