bzoj2819 DFS序 + LCA + 线段树

https://www.lydsy.com/JudgeOnline/problem.php?id=2819

题意：树上单点修改及区间异或和查询。

思维难度不高，但是题比较硬核。

整体思路是维护每一个结点到根节点的距离。查询u，v树链上的异或和就是query(v) ^ query(u) ^ a[lca(u,v)],所以就要想办法维护树上的结点到根节点的异或和。

网上的题解大多是选择直接维护答案，修改的时候修改整颗子树的答案，用线段树或者树状数组区间异或一下l到r内的答案就可以。

我一开始没有想到直接上lca直接维护根节点的距离，所以在dfs序上见了一颗线段树，查询的时候query 1到u第一次出现地方的异或和，区间的时候常规单点修改也可过。

#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
inline int read(){int now=0;register char c=getchar();for(;!isdigit(c);c=getchar());
for(;isdigit(c);now=now*10+c-'0',c=getchar());return now;}
#define For(i, x, y) for(int i=x;i<=y;i++)  
#define _For(i, x, y) for(int i=x;i>=y;i--)
#define Mem(f, x) memset(f,x,sizeof(f))  
#define Sca(x) scanf("%d", &x)
#define Sca2(x,y) scanf("%d%d",&x,&y)
#define Sca3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define Scl(x) scanf("%lld",&x);  
#define Pri(x) printf("%d
", x)
#define Prl(x) printf("%lld
",x);  
#define CLR(u) for(int i=0;i<=N;i++)u[i].clear();
#define LL long long
#define ULL unsigned long long  
#define mp make_pair
#define PII pair<int,int>
#define PIL pair<int,long long>
#define PLL pair<long long,long long>
#define pb push_back
#define fi first
#define se second 
typedef vector<int> VI;
const double eps = 1e-9;
const int maxn = 5e5 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7; 
int N,M,K;
LL a[maxn];
struct Edge{
    int next,to;
}edge[maxn * 2];
int head[maxn],tot;
void init(){
    for(int i = 1; i <= N ; i ++) head[i] = -1;
    tot = 0;
}
void add(int u,int v){
    edge[tot].to = v;
    edge[tot].next = head[u];
    head[u] = tot++;
}
const int SP = 20;
int pa[maxn][SP],dep[maxn];
//求dfs序 
int num;
int dfs_index[maxn * 2];
PII pos[maxn];
void dfs(int t,int la){
    pa[t][0] = la;
    dep[t] = dep[la] + 1;
    for(int i = 1; i <= SP - 1; i ++) pa[t][i] = pa[pa[t][i - 1]][i - 1];
    pos[t].fi = ++num;
    dfs_index[num] = t;
    for(int i = head[t]; ~i ; i = edge[i].next){
        int v = edge[i].to;
        if(v == la) continue;
        dfs(v,t);
    }
    pos[t].se = ++num;
    dfs_index[num] = t;
}
int lca(int u,int v){
    if(dep[u] < dep[v]) swap(u,v);
    int t = dep[u] - dep[v];
    for(int i = 0 ; i <= SP - 1; i ++){
        if(t & (1 << i)) u = pa[u][i];
    }
    for(int i = SP - 1; i >= 0 ; i --){
        int uu = pa[u][i],vv = pa[v][i];
        if(uu != vv){
            u = uu;
            v = vv;
        }
    }
    return u == v? u :pa[u][0];
}
//线段树 
struct Tree{
    LL sum;
}tree[maxn << 3];
void Pushup(int t){
    tree[t].sum = tree[t << 1].sum ^ tree[t << 1 | 1].sum; 
}
void Build(int t,int l,int r){
    if(l == r){
        tree[t].sum = a[dfs_index[l]];
        return;
    }
    int m = (l + r) >> 1;
    Build(t << 1,l,m); Build(t << 1 | 1,m + 1,r);
    Pushup(t);
}
void update(int t,int p,int x,int L,int R){
    if(L == R){
        tree[t].sum = x;
        return;
    }
    int m = (L + R) >> 1;
    if(p <= m) update(t << 1,p,x,L,m);
    else update(t << 1 | 1,p,x,m + 1,R);
    Pushup(t);
}
LL query(int t,int l,int r,int L,int R){
    if(l <= L && R <= r){
        return tree[t].sum;
    }
    int m = (L + R) >> 1;
    if(r <= m) return query(t << 1,l,r,L,m);
    else if(l > m) return query(t << 1 | 1,l,r,m + 1,R);
    else return query(t << 1,l,m,L,m) ^ query(t << 1 | 1,m + 1,r,m + 1,R);
}

LL query(int x){
    return query(1,1,pos[x].fi,1,2 * N);
}
int main(){
    Sca(N); init();
    For(i,1,N) Scl(a[i]);
    For(i,1,N - 1){
        int u,v; Sca2(u,v);
        add(u,v); add(v,u);
    }
    dfs(1,0);
    Build(1,1,2 * N);
    Sca(K);
    while(K--){
        char op[3];
        int u,v;
        scanf("%s%d%d",op,&u,&v);
        if(op[0] == 'Q'){
            if(query(u) ^ query(v) ^ a[lca(u,v)]) puts("Yes");
            else puts("No");
        }else{
            update(1,pos[u].fi,v,1,2 * N);
            update(1,pos[u].se,v,1,2 * N);
            a[u] = v;
        }
    }
    return 0;
}