题目链接：http://codeforces.com/problemset/problem/766/E
大意，给出一个$n$个点的树，每个点都有一个权值，令$Disx$为$u$到$v$路径上的异或和求：
$${sum _{i=1}^{n-1}sum _{j=i}^{n}Disx(i,j)}$$
枚举我当前处理的是这个二进制位上第$k$位的值,就只需要统计每个点的子树中有多少个点到当前点的第$k$位的异或和为$0$，为$1$的分别有多少个，统计答案即可。
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<vector> 5 #include<cstdlib> 6 #include<cmath> 7 #include<cstring> 8 using namespace std; 9 #define maxn 1000100 10 #define llg long long 11 #define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout); 12 llg n,m,k,wei,val[maxn],ans; 13 vector<llg>a[maxn]; 14 15 struct node 16 { 17 llg v1,v2; 18 }; 19 20 void init() 21 { 22 cin>>n; 23 for (llg i=1;i<=n;i++) scanf("%lld",&val[i]),ans+=val[i]; 24 llg x,y; 25 for (llg i=1;i<n;i++) 26 { 27 scanf("%lld%lld",&x,&y); 28 a[x].push_back(y),a[y].push_back(x); 29 } 30 } 31 32 node dfs(llg x,llg fa) 33 { 34 llg w=a[x].size(),v; 35 node now; 36 now.v1=now.v2=0; 37 if (val[x]&wei) now.v1=1;else now.v2=1; 38 for (llg i=0;i<w;i++) 39 { 40 v=a[x][i]; 41 if (v==fa) continue; 42 node ne=dfs(v,x); 43 ans+=now.v1*ne.v2*wei,ans+=now.v2*ne.v1*wei; 44 if (val[x]&wei) now.v1+=ne.v2,now.v2+=ne.v1; else now.v1+=ne.v1,now.v2+=ne.v2; 45 } 46 return now; 47 } 48 49 int main() 50 { 51 yyj("E"); 52 init(); 53 for (k=0;k<=20;k++) 54 { 55 wei=(1<<k); 56 dfs(1,-1); 57 } 58 cout<<ans; 59 return 0; 60 }