bzoj2337

这句话感觉都能成定理了：
xor问题逐位考虑……
这道题就是这样，然后和bzoj3143和相似
但这道题多了自环，于是我们设f[i]表示当前位由i走到n的后为1的数学期望
显然f[n]=0，可得f[i]=sigma((1/d[i])*f[j])（如果边权这位为0）+sigma（(1/d[i])*(1-f[j]))（边权这位为1）
然后高斯消元即可

 type node=record
        po,len,next:longint;
      end;

 var w:array[..] of node;
     b,d,p:array[..] of longint;
     a:array[..,..] of extended;
     n,m,x,y,z,len,t,i,j,k:longint;
     c,ans:extended;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure swap(var a,b:extended);
   var c:extended;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure add(x,y,z:longint);
   begin
     inc(len);
     w[len].po:=y;
     w[len].len:=z;
     w[len].next:=p[x];
     p[x]:=len;
   end;

 procedure work;
   var i,j,k,p:longint;
   begin
     for i:= to n- do
     begin
       p:=i;
       for k:=i+ to n- do
         if abs(a[k,i])>abs(a[p,i]) then p:=k;
       if p<>i then
       begin
         for j:=i to n+ do
           swap(a[p,j],a[i,j]);
       end;
       for k:=i+ to n- do
         if abs(a[k,i])> then
         begin
           for j:=n+ downto i do
             a[k,j]:=a[k,j]-a[i,j]*a[k,i]/a[i,i];
         end;
     end;
     a[n,n+]:=;
     for i:=n- downto  do
     begin
       for j:=i+ to n- do
         a[i,n+]:=a[i,n+]-a[j,n+]*a[i,j];
       a[i,n+]:=a[i,n+]/a[i,i];
     end;
   end;

 begin
   readln(n,m);
   for i:= to m do
   begin
     readln(x,y,z);
     if z<> then t:=max(t,trunc(ln(z)/ln()));
     inc(d[x]);
     inc(d[y]);
     if x=y then dec(d[x])
     else add(y,x,z);
     add(x,y,z);
   end;
   for i:= to t do
   begin
     fillchar(a,sizeof(a),);
     for j:= to n do
       a[j,j]:=;
     for k:= to n do
     begin
       j:=p[k];
       while j<> do
       begin
         y:=w[j].po;
         if w[j].len and ( shl i)= then a[k,y]:=a[k,y]-/d[k]
         else begin
           a[k,y]:=a[k,y]+/d[k];
           a[k,n+]:=a[k,n+]+/d[k];
         end;
         j:=w[j].next;
       end;
     end;
     work;
     ans:=ans+ shl i*a[,n+];
   end;
   writeln(ans::);
 end.