【BZOJ4398】福慧双修（二进制，最短路）

题意：

此题中S=1

思路：Orz ManGod秒切此题

我觉得出入边权互换不太直观，就改了一下写法

第一次默认与1有关的第一条出边只出不入，第二次默认只入不出

var q,dis:array[1..150000]of longint;
    head,vet,next,len,flag,x,y,z,w,id,a,b,c,d:array[1..300000]of longint;
    inq:array[1..50000]of boolean;
    n,m,bz,t3,t4,tot,h1,t1,i,j,ans,m1,t,tmp:longint;

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

function who(x,y:longint):longint;
begin
 if x=1 then exit(y);
 exit(x);
end;

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

procedure spfa;
var u,e,v,t,w:longint;
begin
 t:=0; w:=t1;
 while h1<t1 do
 begin
  inc(h1); inc(t);
  if t=3*n+1 then t:=1;
  u:=q[t]; inq[u]:=false;
  e:=head[u];
  while e<>0 do
  begin
   v:=vet[e];
   if (flag[e]=0)and(dis[u]+len[e]<dis[v]) then
   begin
    dis[v]:=dis[u]+len[e];
    if not inq[v] then
    begin
     inc(t1); inc(w);
     if w=3*n+1 then w:=1;
     q[w]:=v; inq[v]:=true;
    end;
   end;
   e:=next[e];
  end;
 end;
end;

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

begin
 assign(input,'bzoj4398.in'); reset(input);
 assign(output,'bzoj4398.out'); rewrite(output);
 readln(n,m1);
 for i:=1 to m1 do
 begin
  read(x[i],y[i],z[i],w[i]);
  if (x[i]=1)or(y[i]=1) then
  begin
   inc(m); id[m]:=i;
   a[m]:=x[i]; b[m]:=y[i]; c[m]:=z[i]; d[m]:=w[i];
  end;
  add(x[i],y[i],z[i]);
  add(y[i],x[i],w[i]);
 end;

 t:=trunc(ln(n)/ln(2))+1;
 ans:=maxlongint;
 for i:=1 to t do
 begin
  for j:=1 to tot do flag[j]:=0;
  bz:=who(a[1],b[1]);
  if a[1]=1 then flag[id[1]<<1]:=1
   else flag[(id[1]<<1)-1]:=1; //out

  fillchar(inq,sizeof(inq),false);
  fillchar(dis,sizeof(dis),$7f);
  h1:=0; t1:=0;
  inc(t1); q[t1]:=bz; inq[bz]:=true;
  if a[1]=1 then dis[bz]:=c[1]
   else dis[bz]:=d[1];

  for j:=2 to m do
  begin
   tmp:=who(a[j],b[j]);
   t3:=bz and (1<<(i-1));
   t4:=tmp and (1<<(i-1));
   if t3<>t4 then    //in
   begin
    if a[j]=1 then flag[(id[j]<<1)-1]:=1
     else flag[id[j]<<1]:=1;
   end
    else    //out
    begin
     if a[j]=1 then flag[id[j]<<1]:=1
      else flag[(id[j]<<1)-1]:=1;
     inc(t1); q[t1]:=tmp; inq[tmp]:=true;
     if a[j]=1 then dis[tmp]:=c[j]
      else dis[tmp]:=d[j];
    end;
  end;
  spfa;
  ans:=min(ans,dis[1]);
 end;


 for i:=1 to t do
 begin
  for j:=1 to tot do flag[j]:=0;
  bz:=who(a[1],b[1]);
  if a[1]=1 then flag[(id[1]<<1)-1]:=1
   else flag[id[1]<<1]:=1; //in

  fillchar(inq,sizeof(inq),false);
  fillchar(dis,sizeof(dis),$7f);
  h1:=0; t1:=0;


  for j:=2 to m do
  begin
   tmp:=who(a[j],b[j]);
   t3:=bz and (1<<(i-1));
   t4:=tmp and (1<<(i-1));
   if t3<>t4 then    //out
   begin
    if a[j]=1 then flag[id[j]<<1]:=1
     else flag[(id[j]<<1)-1]:=1;
    inc(t1); q[t1]:=tmp; inq[tmp]:=true;
    if a[j]=1 then dis[tmp]:=c[j]
     else dis[tmp]:=d[j];
   end
    else    //in
    begin
     if a[j]=1 then flag[(id[j]<<1)-1]:=1
      else flag[id[j]<<1]:=1;
    end;
  end;
  spfa;
  ans:=min(ans,dis[1]);
 end;
 if ans>1000000000 then writeln(-1)
  else writeln(ans);
 close(input);
 close(output);
end.