poj3411

{二维spfa}

program sky;{由于编辑器导致的格式问题请多包涵}
const
   maxn    = 100000;
type
   rec = record
        y,w,ww,z,next : longint;
     end;
var
   x,y,z,w,ww,n,m : longint;
   i,e,tp,tpp      : longint;
   bian,jia,h,t      : longint;
   q          : array[0..maxn+1,1..2] of longint;
   v          : array[0..11,0..1025] of boolean;
   a          : array[0..100] of rec;
   f          : array[0..11] of longint;
   dis          : array[0..11,0..1025] of longint;
   ans          : longint;

function pd(aa,bb :longint ):boolean;{第一次打位运算处理状态，判断是否走过bb结点}
begin
   if (1<<(aa-1) and bb)<>0 then exit(true);
   exit(false);
end; { pd }
function change(aa,bb :longint ):longint;{走过a[e].y这个结点}
begin
   change:=1<<(aa-1) or bb;
end;
procedure spfa;
begin
   fillchar(dis,sizeof(dis),63);
   dis[1,1]:=0;
   h:=1; t:=1;
   q[1,1]:=1; q[1,2]:=1; v[1,1]:=true;
   while h<>t+1 do
   begin
      e:=f[q[h,1]]; 
      tp:=q[h,1]; tpp:=q[h,2];
      inc(h);if h=maxn then h:=1;
      while e<>0 do
      begin
     if pd(a[e].z,tpp) then{打的有点小麻烦了，bian不用放在里面}
     begin
        bian:=change(a[e].y,tpp);
        jia:=a[e].w;
     end else
     begin
        bian:=change(a[e].y,tpp);{change是加上a[e].y这个点，不是a[e].z......,并且两种情况都要变}
        jia:=a[e].ww;
     end;
     if dis[tp,tpp]+jia<dis[a[e].y,bian] then{仿照一维的spfa，把状态弄进去}
     begin
        dis[a[e].y,bian]:=dis[tp,tpp]+jia;
        if not v[a[e].y,bian] then{需要改变的不仅有dis而且有v，q}
        begin
           v[a[e].y,bian]:=true;
           inc(t); if t=maxn then t:=1;
           q[t,1]:=a[e].y; q[t,2]:=bian;{1存结点，2存状态，用2进制表示}
        end;
     end;
     e:=a[e].next;
      end;
      v[tp,tpp]:=false;
   end;
end;
begin
   readln(n,m);
   for i:=1 to m do
   begin
      read(x,y,z,w,ww);
      inc(e);
      a[e].y:=y; a[e].z:=z;
      a[e].w:=w; a[e].ww:=ww; a[e].next:=f[x];
      f[x]:=e;
   end;
   spfa;
   ans:=maxlongint>>2;
   for i:=0 to 1<<n-1 do{0到1<<n-1为所有状态}
      if ans>dis[n,i] then ans:=dis[n,i];
   if ans=maxlongint>>2 then writeln('impossible'){竟然没看见impossible……这种情况}
      else writeln(ans);{也注意赋的初值，maxlongint>>1是比fillchar63大的，所以div4来判断是否未被更新}
end.

skysun原创。转载请注明出处。http://www.cnblogs.com/skysun