[Code source daily question Div1] Dis "Doubling the LCA maintenance tree chain XOR sum"

Dis

Title description:

Given a tree of n points, each point has its own point weight, ask the XOR sum of the point set formed by the simple path of two points multiple times.

Ideas:

Doubling on the tree at a glance to find the XOR of the LCA maintenance path

fa[i][j]Indicates the 2 ^jth point igoing up from the beginning

fuck[i][j]Represents the XOR sum of the point weights of 2 ⁱi points from the beginning

The process of maintaining the fuck array can be maintained in the process of seeking LCA

It should be noted that when running LCA

• If the sum is in the same chain, the answer at the end xis not counted, we need to judge and calculateyx=y
• If it is not in the same chain, when running LCA, x and y will end up at the next layer node of LCA, which is x-lca-ya situation formed. We have not calculated the values ​​of these three, and we need to calculate them specially.

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define inf 0x3f3f3f3f
#define mod7 1000000007
#define mod9 998244353
#define m_p(a,b) make_pair(a, b)
#define mem(a,b) memset((a),(b),sizeof(a))
#define io ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
typedef long long ll;
typedef pair <int,int> pii;

#define MAX 1000000 + 50

int a, b;
int n, m;
int ar[MAX];
int tot;
int head[MAX];
struct ran{
    
    
    int to, nex;
}tr[MAX];
void add(int u, int v){
    
    
    tr[++tot].to = v;
    tr[tot].nex = head[u];
    head[u] = tot;
}

int deap[MAX];
int fa[MAX][22];
int fuck[MAX][22];
void dfs(int u, int last){
    
    
    deap[u] = deap[last] + 1;
    fa[u][0] = last;
    fuck[u][0] = ar[u];
    for(int i = 1; (1 << i) <= deap[u]; ++i){
    
    
        fa[u][i] = fa[fa[u][i - 1]][i - 1];
        fuck[u][i] = (fuck[u][i-1]^fuck[fa[u][i-1]][i-1]);
    }
    for(int i = head[u]; i; i = tr[i].nex){
    
    
        int v = tr[i].to;
        if(v != last)dfs(v, u);
    }
}

int lca(int x, int y){
    
    
    if(deap[x] < deap[y])swap(x, y);
    int ans = 0;
    for(int i = 20; i >= 0; --i){
    
    
        if(deap[fa[x][i]] >= deap[y]){
    
    
            ans ^= fuck[x][i];
            x = fa[x][i];
        }
    }
    if(x == y){
    
    
        ans ^= ar[x];
        return ans;
    }
    for(int i = 20; i >= 0; --i){
    
    
        if(fa[x][i] != fa[y][i]){
    
    
            ans ^= fuck[x][i];
            ans ^= fuck[y][i];
            x = fa[x][i];
            y = fa[y][i];
        }
    }
    ans ^= fuck[x][1] ^ fuck[y][1] ^ ar[fa[x][0]];
    return ans;
}


void work(){
    
    
    cin >> n >> m;
    for(int i = 1; i <= n; ++i)cin >> ar[i];
    for(int i = 1; i < n; ++i){
    
    
        cin >> a >> b;
        add(a, b);
        add(b, a);
    }
    dfs(1, 0);
    for(int i = 1; i <= m; ++i){
    
    
        cin >> a >> b;
        cout<<lca(a, b)<<endl;
    }
}


int main(){
    
    
    io;
    work();
    return 0;
}