2337:[HNOI2011]XOR和路径 - BZOJ

昨天才做了一道高斯消元，一下要精度判断，一下又不要精度判断
主要是思路很重要
很容易想到每一个二进制位算一个概率，然后求和，设f[i]为走到从i走到n这一个二进制位为1的概率
f[i]:=∑{f[j]/d[i](i到j的路径这一位是0),(1-f[j])/d[i],(i到j的路径这一位是1)}(f[n]直接设为0)

 var
     f:array[..,..]of extended;
     d:array[..]of longint;
     v,u,w:array[..]of longint;
     ans:extended;
     n,m:longint;

 procedure swap(var x,y:extended);
 var
     t:extended;
 begin
     t:=x;
     x:=y;
     y:=t;
 end;

 procedure init;
 var
     i:longint;
 begin
     read(n,m);
     for i:= to m do
       begin
         read(v[i],u[i],w[i]);
         inc(d[v[i]]);
         inc(d[u[i]]);
         if u[i]=v[i] then dec(d[u[i]]);
       end;
 end;

 procedure work;
 var
     t,i,j,k:longint;
     s:extended;
 begin
     for t:= to  do
       begin
         for i:= to n do
           for j:= to n+ do
             f[i,j]:=;
         for i:= to n do
           f[i,i]:=;
         for i:= to m do
           if w[i]and(<<t)= then
             begin
               f[v[i],u[i]]:=f[v[i],u[i]]-/d[v[i]];
               f[u[i],v[i]]:=f[u[i],v[i]]-/d[u[i]];
               if u[i]=v[i] then f[v[i],v[i]]:=f[u[i],u[i]]+/d[v[i]];
             end
           else
             begin
               f[v[i],u[i]]:=f[v[i],u[i]]+/d[v[i]];
               f[u[i],v[i]]:=f[u[i],v[i]]+/d[u[i]];
               f[v[i],n+]:=f[v[i],n+]+/d[v[i]];
               f[u[i],n+]:=f[u[i],n+]+/d[u[i]];
               if v[i]=u[i] then
               begin
                 f[u[i],v[i]]:=f[u[i],v[i]]-/d[u[i]];
                 f[v[i],n+]:=f[v[i],n+]-/d[v[i]];
               end;
             end;
         for i:= to n- do
           begin
             for j:=i to n- do
               if f[j,i]<> then break;
             for k:=i to n+ do
               swap(f[i,k],f[j,k]);
             for j:=i+ to n- do
               if abs(f[j,i])> then
               begin
                 s:=f[j,i]/f[i,i];
                 f[j,i]:=;
                 for k:=i+ to n+ do
                   f[j,k]:=f[j,k]-s*f[i,k];
               end;
           end;
           for i:=n- downto  do
             begin
               for j:=i+ to n- do
                 f[i,n+]:=f[i,n+]-f[j,]*f[i,j];
               f[i,]:=f[i,n+]/f[i,i];
             end;
           ans:=ans+f[,]*(<<t);
       end;
     write(ans::);
 end;