第四届传智杯 F1(还就内阁看错题)

题目
题意: 给定n个点的树，规定两点之间的路径为异或和。给定m组询问，判断是否存在一点t，使得 dis(t,a) ^ dis(t,b) = c.
思路: 看错题了，以为路径是和呢。如果是异或和那纯纯简单了。因为a、b路径外的点引的边二者都会异或到，相当于没有。如果a、b路径内的和以a、b为起点一样的。
时间复杂度: O(n+m)
代码:

// Problem: 距离
// Contest: AcWing
// URL: https://www.acwing.com/problem/content/description/1173/
// Memory Limit: 64 MB
// Time Limit: 1000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 5e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
int n,m,k,T;
#define int long long
int h[N],e[N<<1],ne[N<<1],w[N<<1],idx;
int dist[N];
void add(int a,int b,int c)
{
	e[idx] = b,w[idx] = c,ne[idx] = h[a],h[a] = idx++;
}
void dfs(int cur,int father)
{
	for(int i=h[cur];~i;i=ne[i])
	{
		int j = e[i];
		if(j == father) continue;
		dist[j] = dist[cur] ^ w[i];
		dfs(j,cur);
	}
}
void solve()
{
   mem(h,-1); idx = 0;
   read(n); read(m);
   for(int i=0;i<n-1;++i)
   {
   	 int x,y,z; read(x),read(y),read(z);
   	 add(x,y,z),add(y,x,z);
   }
   dfs(1,0);
   while(m--)
   {
   	 int x,y; read(x),read(y); int z; read(z);
   	 if(z == (dist[x]^dist[y]))
   	 {
   	 	 puts("YES");
   	 }
   	 else puts("NO");
   }
}
signed main(void)
{  
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}