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=9999999999;

var map,pm:array[0..110,0..110] of int64;
    f,pf:array[0..110] of int64;
    v:array[0..1010] 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]=0 then
    begin
      inc(t);
      v[x]:=t;
    end;
    exit(v[x]);
  end;

procedure time1;
  var i,j:longint;
  begin
    pf:=f;
    for i:=1 to t do
    begin
      f[i]:=inf;
      for j:=1 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:=1 to t do
      for j:=1 to t do
      begin
        map[i,j]:=inf;
        for k:=1 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>0 do
    begin
      if x mod 2=1 then time1;
      x:=x shr 1;
      if x<>0 then time2;
    end;
  end;

begin
  readln(n,m,s,e);
  for i:=1 to 100 do
    for j:=1 to 100 do
      map[i,j]:=inf;
  for i:=1 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:=1 to t do
    f[i]:=inf;
  f[s]:=0;
  work(n);
  writeln(f[e]);
end.