SPOJ10707 COT2 - Count on a tree II 【树上莫队】

题目分析：

考虑欧拉序，这里的欧拉序与ETT欧拉序的定义相同而与倍增LCA不同。然后不妨对于询问$u$与$v$让$dfsin[u] \leq dfsin[v]$，这样对于u和v不在一条路径上，它们可以改成询问$dfsin[u]$到$dfsin[v]$。否则改成$dfsout[u]$到$dfsin[v]$，并加上LCA上的影响，如果在询问过程中没有加入就加入，否则删除。

代码：

 #include<bits/stdc++.h>
 using namespace std;
 
 const int maxn = ;
 const int maxm = ;
 const int srt = ;
 
 int n,m;
 int a[maxn],tb[maxn],arr[maxn],fa[maxn],pm[maxm],ans[maxm];
 int dfsin[maxn],dfsout[maxn],Num,pre[maxn];
 vector <int> g[maxn];
 vector <pair<int,int> > Qy[maxn];
 struct query{int l,r,bel,rem;}Q[maxm];
 
 int found(int x){
     int rx = x; while(pre[rx] != rx) rx = pre[rx];
     while(pre[x] != rx){int tmp = pre[x]; pre[x] = rx; x = tmp;}
     return rx;
 }
 
 void dfs(int now,int ff){
     arr[now] = ;pm[++Num] = now;fa[now] = ff;
     dfsin[now] = Num;
     for(int i=;i<Qy[now].size();i++){
     if(!arr[Qy[now][i].first])continue;
     Q[Qy[now][i].second].bel=found(Qy[now][i].first);
     }
     for(int i=;i<g[now].size();i++){
     if(g[now][i]==fa[now])continue;
     dfs(g[now][i],now);
     }
     pre[now] = fa[now];
     pm[++Num] = now; dfsout[now] = Num;
 }
 
 void read(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++) scanf("%d",&a[i]),tb[i] = a[i];
     sort(tb+,tb+n+);  int num=unique(tb+,tb+n+)-tb-;
     for(int i=;i<=n;i++) a[i]=lower_bound(tb+,tb+num+,a[i])-tb;
     memset(tb,,sizeof(tb));
     for(int i=;i<n;i++){
     int u,v; scanf("%d%d",&u,&v);
     g[u].push_back(v); g[v].push_back(u);
     }
     for(int i=;i<=m;i++) scanf("%d%d",&Q[i].l,&Q[i].r);
 }
 
 int cmp(query alpha,query beta){
     if(alpha.l/srt == beta.l/srt) return alpha.r < beta.r;
     else return alpha.l/srt < beta.l/srt;
 }
 
 void init(){
     for(int i=;i<=m;i++){
     Qy[Q[i].l].push_back(make_pair(Q[i].r,i));
     Qy[Q[i].r].push_back(make_pair(Q[i].l,i));
     }
     for(int i=;i<=n;i++) pre[i] = i;
     dfs(,);
     for(int i=;i<=m;i++){
     if(dfsin[Q[i].l] > dfsin[Q[i].r]) swap(Q[i].l,Q[i].r);
     if(Q[i].bel == Q[i].l || Q[i].bel == Q[i].r){
         Q[i].l = dfsin[Q[i].l]; Q[i].r = dfsin[Q[i].r];
         Q[i].bel = i; Q[i].rem = ;
     }else{
         Q[i].l = dfsout[Q[i].l]; Q[i].r = dfsin[Q[i].r];
         Q[i].rem = Q[i].bel; Q[i].bel = i;
     }
     }
     sort(Q+,Q+m+,cmp);
 }
 
 void work(){
     int L = ,R = ,as=; tb[a[]]++;
     for(int i=;i<=m;i++){
     int nowl = Q[i].l,nowr = Q[i].r;
     while(R < nowr){
         R++;
         if(dfsin[pm[R]]==R||dfsin[pm[R]]<L){
         if(!tb[a[pm[R]]])as++;
         tb[a[pm[R]]]++;
         }else{
         if(tb[a[pm[R]]] == ) as--;
         tb[a[pm[R]]]--;
         }
     }
     while(L > nowl){
         L--;
         if(dfsout[pm[L]]==L||dfsout[pm[L]]>R){
         if(!tb[a[pm[L]]])as++;
         tb[a[pm[L]]]++;
         }else{
         if(tb[a[pm[L]]] == ) as--;
         tb[a[pm[L]]]--;
         }
     }
     while(R > nowr){
         if(dfsin[pm[R]]==R||dfsin[pm[R]]<L){
         if(tb[a[pm[R]]] == ) as--;
         tb[a[pm[R]]]--;
         }else{
         if(!tb[a[pm[R]]])as++;
         tb[a[pm[R]]]++;
         }
         R--;
     }
     while(L < nowl){
         if(dfsout[pm[L]]==L||dfsout[pm[L]]>R){
         if(tb[a[pm[L]]] == ) as--;
         tb[a[pm[L]]]--;
         }else{
         if(!tb[a[pm[L]]])as++;
         tb[a[pm[L]]]++;
         }
         L++;
     }
     if(Q[i].rem && !tb[a[Q[i].rem]])ans[Q[i].bel]=as+;
     else ans[Q[i].bel] = as;
     }
     for(int i=;i<=m;i++) printf("%d\n",ans[i]);
 }
 
 int main(){
     read();
     init();
     work();
     return ;
 }