BZOJ5338 [TJOI2018] Xor 【可持久化Trie树】【dfs序】

题目分析：

　　很无聊的一道题目。首先区间内单点对应异或值的询问容易想到trie树。由于题目在树上进行，case1将路径分成两段，然后dfs的时候顺便可持久化trie树做询问。case2维护dfs序，对dfs序建可持久化的trie树。这样做的空间复杂度是O(nw)，时间复杂度是O(nw).

代码：

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

const int maxn=102000;

int n,q;
int v[maxn];
vector<int> g[maxn];

vector<pair<int,int> > qy[maxn];
vector<pair<int,int> > vec;

int ans[maxn],num,kd[maxn];
int dep[maxn],dfsin[maxn],dfsout[maxn],fa[maxn];

vector<pair<int,int> > Lca[maxn];
int pre[maxn];

int found(int x){
    int rx = x; while(pre[rx] != rx) rx = pre[rx];
    while(pre[x] != rx){
    int t = pre[x]; pre[x] = rx; x = t;
    }
    return rx;
}

void dfs(int now,int dp,int f){
    dfsin[now] = dfsout[now] = ++num;fa[now] = f; dep[now] = dp;
    for(auto pr:Lca[now]){
    if(dep[pr.first] == 0) continue;
    int la = found(pr.first);
    qy[now].push_back(make_pair(la,pr.second));
    qy[pr.first].push_back(make_pair(la,pr.second));
    }
    for(auto i:g[now]){
    if(i == f) continue;
    dfs(i,dp+1,now);
    dfsout[now] = dfsout[i];
    }
    pre[found(now)] = found(f);
}

void read(){
    scanf("%d%d",&n,&q);
    for(int i=1;i<=n;i++) scanf("%d",&v[i]);
    for(int i=1;i<n;i++){
    int x,y; scanf("%d%d",&x,&y);
    g[x].push_back(y); g[y].push_back(x);
    }
    for(int i=1;i<=q;i++){
    int cas; scanf("%d",&cas);
    if(cas == 1){
        int x,y; scanf("%d%d",&x,&y);
        vec.push_back(make_pair(x,i));kd[i] = y;
    }else{
        int x,y,z; scanf("%d%d%d",&x,&y,&z);
        Lca[x].push_back(make_pair(y,i));
        if(x!=y) Lca[y].push_back(make_pair(x,i));
        kd[i] = z;
    }
    }
    for(int i=1;i<=n;i++) pre[i] = i;
    dfs(1,1,0);
}

int his[maxn],om;
int sz[maxn*130],ch[maxn*130][2];

void follow(int pt,int last,int dt){
    int hbit = 30;
    while(hbit!=-1){
    if((1<<hbit)&dt){
        ch[pt][0] = ch[last][0];
        ch[pt][1] = ++num;
        pt = num; last = ch[last][1];
        sz[pt] = sz[last]+1;
    }else{
        ch[pt][1] = ch[last][1];
        ch[pt][0] = ++num;
        pt = num; last = ch[last][0];
        sz[pt] = sz[last]+1;
    }
    hbit--;
    }
}

void ins(int now){
    int last = his[om-1];
    num++;int pt = num; sz[pt] = sz[last]+1;
    his[om] = num;
    follow(pt,last,v[now]);
}

int query(int now,int last,int z){
    now = his[now]; last = his[last];
    int hbit = 30;
    while(hbit!=-1){
    if((1<<hbit)&z){
        if(sz[ch[now][0]]-sz[ch[last][0]]){
        now = ch[now][0];last = ch[last][0];
        }else{
        z -= (1<<hbit);now = ch[now][1];last = ch[last][1];
        }
    }else{
        if(sz[ch[now][1]]-sz[ch[last][1]]){
        z += (1<<hbit);
        now = ch[now][1];last = ch[last][1];
        }else{
        now = ch[now][0];last = ch[last][0];
        }
    }
    hbit--;
    }
    return z;
}

void del(int now){his[om] = 0;}

void dfs2(int now){
    om++;ins(now);
    for(auto pr:qy[now]){
    int last = dep[pr.first]-1,data = kd[pr.second];
    ans[pr.second] = max(ans[pr.second],query(dep[now],last,data));
    }
    for(auto i:g[now]){
    if(i == fa[now]) continue;
    dfs2(i); 
    }
    del(now);om--;
}

void dfs3(int now){
    om++;ins(now);
    for(auto i:g[now]){
    if(i == fa[now]) continue;
    dfs3(i);
    }
}

void work(){
    num = 1;his[0] = 1;
    dfs2(1);
    memset(ch,0,sizeof(ch));memset(sz,0,sizeof(sz));
    num = 1;his[0] = 1;om = 0;
    dfs3(1);
    for(auto pr:vec){
    int st = dfsin[pr.first],ed = dfsout[pr.first];
    ans[pr.second] = query(ed,st-1,kd[pr.second]);
    }
    for(int i=1;i<=q;i++) printf("%d
",ans[i]);
}

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