都是简单的异或高斯消元
由于bzoj2466要求解得最小和，所以我们最后还要穷举自由元取最优解
1 type node=record 2 po,next:longint; 3 end; 4 5 var f:array[0..110] of longint; 6 a:array[0..110,0..110] of longint; 7 i,x,y,ans,n:longint; 8 9 procedure swap(var a,b:longint); 10 var c:longint; 11 begin 12 c:=a; 13 a:=b; 14 b:=c; 15 end; 16 17 procedure dfs(k,sum:longint); 18 var j,p:longint; 19 begin 20 if sum>=ans then exit; 21 if k=0 then 22 begin 23 if ans>sum then ans:=sum; 24 exit; 25 end; 26 if a[k,k]=1 then 27 begin 28 p:=a[k,n+1]; 29 for j:=k+1 to n do 30 if a[k,j]=1 then p:=p xor f[j]; //回代 31 f[k]:=p; 32 dfs(k-1,sum+f[k]); 33 end 34 else begin 35 f[k]:=0; 36 dfs(k-1,sum); 37 f[k]:=1; 38 dfs(k-1,sum+1); 39 f[k]:=0; 40 end; 41 end; 42 43 procedure work; 44 var i,j,k,p:longint; 45 begin 46 for i:=1 to n do 47 begin 48 p:=i; 49 for j:=i to n do 50 if a[j,i]<>0 then 51 begin 52 p:=j; 53 break; 54 end; 55 56 if p<>i then 57 for j:=1 to n+1 do 58 swap(a[i,j],a[p,j]); 59 if a[i,i]=0 then continue; 60 for k:=i+1 to n do 61 if a[k,i]<>0 then 62 begin 63 for j:=i to n+1 do 64 a[k,j]:=a[k,j] xor a[i,j]; 65 end; 66 end; 67 ans:=361; 68 dfs(n,0); 69 end; 70 71 begin 72 readln(n); 73 while n<>0 do 74 begin 75 fillchar(a,sizeof(a),0); 76 fillchar(f,sizeof(f),0); 77 for i:=1 to n-1 do 78 begin 79 readln(x,y); 80 a[x,y]:=1; 81 a[y,x]:=1; 82 end; 83 for i:=1 to n do 84 begin 85 a[i,i]:=1; 86 a[i,n+1]:=1; 87 end; 88 work; 89 writeln(ans); 90 readln(n); 91 end; 92 end.
