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
1 type node=record 2 po,len,next:longint; 3 end; 4 5 var w:array[0..20010] of node; 6 b,d,p:array[0..110] of longint; 7 a:array[0..110,0..110] of extended; 8 n,m,x,y,z,len,t,i,j,k:longint; 9 c,ans:extended;10 11 function max(a,b:longint):longint;12 begin13 if a>b then exit(a) else exit(b);14 end;15 16 procedure swap(var a,b:extended);17 var c:extended;18 begin19 c:=a;20 a:=b;21 b:=c;22 end;23 24 procedure add(x,y,z:longint);25 begin26 inc(len);27 w[len].po:=y;28 w[len].len:=z;29 w[len].next:=p[x];30 p[x]:=len;31 end;32 33 procedure work;34 var i,j,k,p:longint;35 begin36 for i:=1 to n-2 do37 begin38 p:=i;39 for k:=i+1 to n-1 do40 if abs(a[k,i])>abs(a[p,i]) then p:=k;41 if p<>i then42 begin43 for j:=i to n+1 do44 swap(a[p,j],a[i,j]);45 end;46 for k:=i+1 to n-1 do47 if abs(a[k,i])>0 then48 begin49 for j:=n+1 downto i do50 a[k,j]:=a[k,j]-a[i,j]*a[k,i]/a[i,i];51 end;52 end;53 a[n,n+1]:=0;54 for i:=n-1 downto 1 do55 begin56 for j:=i+1 to n-1 do57 a[i,n+1]:=a[i,n+1]-a[j,n+1]*a[i,j];58 a[i,n+1]:=a[i,n+1]/a[i,i];59 end;60 end;61 62 begin63 readln(n,m);64 for i:=1 to m do65 begin66 readln(x,y,z);67 if z<>0 then t:=max(t,trunc(ln(z)/ln(2)));68 inc(d[x]);69 inc(d[y]);70 if x=y then dec(d[x])71 else add(y,x,z);72 add(x,y,z);73 end;74 for i:=0 to t do75 begin76 fillchar(a,sizeof(a),0);77 for j:=1 to n do78 a[j,j]:=1;79 for k:=1 to n do80 begin81 j:=p[k];82 while j<>0 do83 begin84 y:=w[j].po;85 if w[j].len and (1 shl i)=0 then a[k,y]:=a[k,y]-1/d[k]86 else begin87 a[k,y]:=a[k,y]+1/d[k];88 a[k,n+1]:=a[k,n+1]+1/d[k];89 end;90 j:=w[j].next;91 end;92 end;93 work;94 ans:=ans+1 shl i*a[1,n+1];95 end;96 writeln(ans:0:3);97 end.