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=;

 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] == ) 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+,now);
     dfsout[now] = dfsout[i];
     }
     pre[found(now)] = found(f);
 }

 void read(){
     scanf("%d%d",&n,&q);
     for(int i=;i<=n;i++) scanf("%d",&v[i]);
     for(int i=;i<n;i++){
     int x,y; scanf("%d%d",&x,&y);
     g[x].push_back(y); g[y].push_back(x);
     }
     for(int i=;i<=q;i++){
     int cas; scanf("%d",&cas);
     if(cas == ){
         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=;i<=n;i++) pre[i] = i;
     dfs(,,);
 }

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

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

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

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

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

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

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