bzoj1706

倍增好题,f[p,i,j]表示i到j经过了2^p条边走过的最短路径
显然f[p+1]可以由f[p]转移来
然后对n二进制拆分累加即可

 const inf=;

 var map,pm:array[..,..] of int64;
     f,pf:array[..] of int64;
     v:array[..] of longint;
     j,z,x,y,s,t,n,m,e,i:longint;

 function min(a,b:int64):int64;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function get(x:longint):longint;
   begin
     if v[x]= then
     begin
       inc(t);
       v[x]:=t;
     end;
     exit(v[x]);
   end;

 procedure time1;
   var i,j:longint;
   begin
     pf:=f;
     for i:= to t do
     begin
       f[i]:=inf;
       for j:= to t do
         f[i]:=min(f[i],pf[j]+map[j,i]);
     end;
   end;

 procedure time2;
   var i,j,k:longint;
   begin
     pm:=map;
     for i:= to t do
       for j:= to t do
       begin
         map[i,j]:=inf;
         for k:= to t do
           map[i,j]:=min(map[i,j],pm[i,k]+pm[k,j]);
       end;
   end;

 procedure work(x:longint);
   begin
     while x> do
     begin
       if x mod = then time1;
       x:=x shr ;
       if x<> then time2;
     end;
   end;

 begin
   readln(n,m,s,e);
   for i:= to  do
     for j:= to  do
       map[i,j]:=inf;
   for i:= to m do
   begin
     readln(z,x,y);
     x:=get(x);
     y:=get(y);
     map[x,y]:=z;
     map[y,x]:=z;
   end;
   s:=get(s);
   e:=get(e);
   for i:= to t do
     f[i]:=inf;
   f[s]:=;
   work(n);
   writeln(f[e]);
 end.