SPOJ10707 COT2

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 = 40200;
const int maxm = 101000;
const int srt = 325;

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] = 1;pm[++Num] = now;fa[now] = ff;
    dfsin[now] = Num;
    for(int i=0;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=0;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=1;i<=n;i++) scanf("%d",&a[i]),tb[i] = a[i];
    sort(tb+1,tb+n+1);  int num=unique(tb+1,tb+n+1)-tb-1;
    for(int i=1;i<=n;i++) a[i]=lower_bound(tb+1,tb+num+1,a[i])-tb;
    memset(tb,0,sizeof(tb));
    for(int i=1;i<n;i++){
    int u,v; scanf("%d%d",&u,&v);
    g[u].push_back(v); g[v].push_back(u);
    }
    for(int i=1;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=1;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=1;i<=n;i++) pre[i] = i;
    dfs(1,0);
    for(int i=1;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 = 0;
    }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+1,Q+m+1,cmp);
}

void work(){
    int L = 1,R = 1,as=1; tb[a[1]]++;
    for(int i=1;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]]] == 1) 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]]] == 1) as--;
        tb[a[pm[L]]]--;
        }
    }
    while(R > nowr){
        if(dfsin[pm[R]]==R||dfsin[pm[R]]<L){
        if(tb[a[pm[R]]] == 1) 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]]] == 1) 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+1;
    else ans[Q[i].bel] = as;
    }
    for(int i=1;i<=m;i++) printf("%d
",ans[i]);
}

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