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)
1 var 2 f:array[0..110,0..110]of extended; 3 d:array[0..110]of longint; 4 v,u,w:array[0..10010]of longint; 5 ans:extended; 6 n,m:longint; 7 8 procedure swap(var x,y:extended); 9 var 10 t:extended; 11 begin 12 t:=x; 13 x:=y; 14 y:=t; 15 end; 16 17 procedure init; 18 var 19 i:longint; 20 begin 21 read(n,m); 22 for i:=1 to m do 23 begin 24 read(v[i],u[i],w[i]); 25 inc(d[v[i]]); 26 inc(d[u[i]]); 27 if u[i]=v[i] then dec(d[u[i]]); 28 end; 29 end; 30 31 procedure work; 32 var 33 t,i,j,k:longint; 34 s:extended; 35 begin 36 for t:=0 to 30 do 37 begin 38 for i:=1 to n do 39 for j:=1 to n+1 do 40 f[i,j]:=0; 41 for i:=1 to n do 42 f[i,i]:=1; 43 for i:=1 to m do 44 if w[i]and(1<<t)=0 then 45 begin 46 f[v[i],u[i]]:=f[v[i],u[i]]-1/d[v[i]]; 47 f[u[i],v[i]]:=f[u[i],v[i]]-1/d[u[i]]; 48 if u[i]=v[i] then f[v[i],v[i]]:=f[u[i],u[i]]+1/d[v[i]]; 49 end 50 else 51 begin 52 f[v[i],u[i]]:=f[v[i],u[i]]+1/d[v[i]]; 53 f[u[i],v[i]]:=f[u[i],v[i]]+1/d[u[i]]; 54 f[v[i],n+1]:=f[v[i],n+1]+1/d[v[i]]; 55 f[u[i],n+1]:=f[u[i],n+1]+1/d[u[i]]; 56 if v[i]=u[i] then 57 begin 58 f[u[i],v[i]]:=f[u[i],v[i]]-1/d[u[i]]; 59 f[v[i],n+1]:=f[v[i],n+1]-1/d[v[i]]; 60 end; 61 end; 62 for i:=1 to n-2 do 63 begin 64 for j:=i to n-1 do 65 if f[j,i]<>0 then break; 66 for k:=i to n+1 do 67 swap(f[i,k],f[j,k]); 68 for j:=i+1 to n-1 do 69 if abs(f[j,i])>0 then 70 begin 71 s:=f[j,i]/f[i,i]; 72 f[j,i]:=0; 73 for k:=i+1 to n+1 do 74 f[j,k]:=f[j,k]-s*f[i,k]; 75 end; 76 end; 77 for i:=n-1 downto 1 do 78 begin 79 for j:=i+1 to n-1 do 80 f[i,n+1]:=f[i,n+1]-f[j,0]*f[i,j]; 81 f[i,0]:=f[i,n+1]/f[i,i]; 82 end; 83 ans:=ans+f[1,0]*(1<<t); 84 end; 85 write(ans:0:3); 86 end;
