codeforces 766E Mahmoud and a xor trip

题目链接：http://codeforces.com/problemset/problem/766/E

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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：

$${\sum _{i=1}^{n-1}\sum _{j=i}^{n}Disx(i,j)}$$

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 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。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstdlib>
#include<cmath>
#include<cstring>
using namespace std;
#define maxn 1000100
#define llg long long 
#define yyj(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
llg n,m,k,wei,val[maxn],ans;
vector<llg>a[maxn];

struct node
{
    llg v1,v2;
};

void init()
{
    cin>>n;
    for (llg i=1;i<=n;i++) scanf("%lld",&val[i]),ans+=val[i];
    llg x,y;
    for (llg i=1;i<n;i++)
    {
        scanf("%lld%lld",&x,&y);
        a[x].push_back(y),a[y].push_back(x);
    }
}

node dfs(llg x,llg fa)
{
    llg w=a[x].size(),v;
    node now;
    now.v1=now.v2=0;
    if (val[x]&wei) now.v1=1;else now.v2=1;
    for (llg i=0;i<w;i++) 
    {
          v=a[x][i];
        if (v==fa) continue;
        node ne=dfs(v,x);
        ans+=now.v1*ne.v2*wei,ans+=now.v2*ne.v1*wei;
        if (val[x]&wei) now.v1+=ne.v2,now.v2+=ne.v1; else now.v1+=ne.v1,now.v2+=ne.v2;     
    }
    return now;
}

int main()
{
    yyj("E");
    init();
    for (k=0;k<=20;k++)
    {
        wei=(1<<k);
        dfs(1,-1);
    }
    cout<<ans;
    return 0;
}