异或的路径

https://ac.nowcoder.com/acm/contest/911/J

思路：dfs处理出根节点到全部节点的异或值，然后对每一位二进制计算贡献；第i位就是2^i的贡献，而贡献的数量就是0和1的数量积，因为只有端点是0和1路径异或后才是1，该位才会1；最好乘二是终起点可以调换；

#include<algorithm>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstdio>
#include<map>
#include<stack>
#include<string>
#include<bits/stdc++.h>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define sfs(i) scanf("%s",(i))
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define ull unsigned long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-16
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
#define MOD(x) ((x%mod)+mod)%mod
#define endl '\n'
#define pb push_back
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FAST_IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)

const int maxn=1e5+9;
const int mod=1e9+7;

inline ll read()
{
    ll f=1,x=0;
    char ss=getchar();
    while(ss<'0'||ss>'9')
    {
        if(ss=='-')f=-1;ss=getchar();
    }
    while(ss>='0'&&ss<='9')
    {
        x=x*10+ss-'0';ss=getchar();
    }    return f*x;
}

vector<pair<int,int> >g[maxn];
int n;
int val[maxn];
void add(int u,int v,int w)
{
    g[u].pb({v,w});
}

void dfs(int u)
{
    for(int i=0;i<g[u].size();i++)
    {
        int v=g[u][i].first;
        int w=g[u][i].second;
        val[v]=val[u]^w;
        dfs(v);
    }
}
int main()
{
    FAST_IO;
    //freopen("input.txt","r",stdin);

    cin>>n;
    for(int i=1;i<=n-1;i++)
    {
        int p;
        int w;
        cin>>p>>w;
        add(p,i+1,w);
    }
    dfs(1);
    ll ans=0;
    for(int i=0;i<=20;i++)
    {
        ll yi=0;
        for(int j=1;j<=n;j++)
        {
            if((val[j]>>i)&1) yi++;
        }
        ans=ans+(yi*(n-yi)%mod)*((1L<<i)%mod);
        ans%=mod;
    }
    cout<<ans*2%mod<<endl;
}