链接
题解
建立圆方树之后，原问题转化为：
询问一个区间里面有多少种值出现了奇数次/偶数次？
有一个经典问题就是询问区间有多少种不同的值，这个可以莫队解决
那这个题其实就几乎一样，直接莫队做就行了
另一种可能的做法
直接 好像就结束了，但是应该代码也不会比我的做法短多少
代码
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
struct Graph
{
int etot, head[maxn], to[maxe], next[maxe], w[maxe];
void clear(int N)
{
for(int i=1;i<=N;i++)head[i]=0;
etot=0;
}
void adde(int a, int b, int c=0){to[++etot]=b;w[etot]=c;next[etot]=head[a];head[a]=etot;}
#define forp(_,__) for(auto p=__.head[_];p;p=__.next[p])
}G;
struct Circle_Square_Tree
{
ll dfn[maxn], low[maxn], tim, tot, s[maxn], e[maxn], n;
vector<ll> cir[maxn], w[maxn];
Graph T;
void dfs(Graph &G, ll u, ll fa)
{
ll ch=0;
s[++*s]=u;
dfn[u]=low[u]=++tim;
forp(u,G)
{
auto v=G.to[p];
if(!dfn[v])
{
ch++;
e[v]=p;
dfs(G,v,u);
low[u]=min(low[u],low[v]);
if(low[v]==dfn[u])
{
if(s[*s]==v)
{
ll W;
forp(u,G)if(G.to[p]==s[*s])W=G.w[p];
T.adde(v,u,W);
T.adde(u,v,W);
--*s;
}
else
{
tot++;
forp(u,G)
if(G.to[p]==s[*s])
w[tot].emb(G.w[p]);
for(ll x=0;x!=v;--*s)
{
x=s[*s];
cir[tot].emb(x);
ll bk=w[tot].back();
w[tot].emb(bk+G.w[e[x]]);
}
cir[tot].emb(u);
ll i; rep(i,0,cir[tot].size()-1)
{
ll W=min((ll)w[tot].at(i),w[tot].back()-w[tot].at(i));
T.adde(n+tot,cir[tot].at(i),W);
T.adde(cir[tot].at(i),n+tot,W);
}
}
}
}
else low[u]=min(low[u],dfn[v]);
}
}
void build(Graph &G, ll N)
{
ll i;
for(i=1;i<=N;i++)dfn[i]=low[i]=0;
T.clear(N*2);
*s=tim=tot=0;
n=N;
for(i=1;i<=N;i++)if(!dfn[i])dfs(G,i,-1);
}
}cstree;
struct BIT
{
ll bit[maxn], n;
void init(int N){n=N;for(int i=1;i<=n;i++)bit[i]=0;}
ll lowbit(ll x){return x&(-x);}
void add(ll pos, ll v)
{
for(;pos<=n;pos+=lowbit(pos))bit[pos]+=v;
}
ll sum(ll pos)
{
ll ans(0);
for(;pos;pos-=lowbit(pos))ans+=bit[pos];
return ans;
}
}bit[2];
struct Easy_Tree
{
int depth[maxn], dist[maxn], tid[maxn], rtid[maxn], tim, size[maxn], rev[maxn];
void dfs(int pos, int pre, Graph& G)
{
tid[pos]=++tim;
rev[tid[pos]]=pos;
size[pos]=1;
forp(pos,G)if(G.to[p]!=pre)
{
depth[G.to[p]]=depth[pos]+1;
dist[G.to[p]]=dist[pos]+G.w[p];
dfs(G.to[p],pos,G);
size[pos]+=size[G.to[p]];
}
rtid[pos]=tim;
}
void run(Graph& G, int root)
{
tim=0;
depth[root]=1;
dfs(1,0,G);
}
}et;
ll cnt[maxn], a[maxn], id[maxn], y[maxn], type[maxn], l[maxn], r[maxn], ans[maxn];
void upd(ll x, ll opt)
{
x=a[et.rev[x]];
if(x==0)return;
if(cnt[x])bit[cnt[x]&1].add(x,-1);
cnt[x]+=opt;
if(cnt[x])bit[cnt[x]&1].add(x,+1);
}
int main()
{
ll i, n, q, u, v, m, L, R;
n=read(), m=read();
rep(i,1,n)a[i]=read();
rep(i,1,m)
{
u=read(), v=read();
G.adde(u,v), G.adde(v,u);
}
cstree.build(G,n);
et.run(cstree.T,1);
q=read();
rep(i,1,q)
{
type[i]=read();
ll x=read();
l[i]=et.tid[x];
r[i]=et.rtid[x];
y[i]=read();
id[i]=i;
}
ll S=sqrt(1e5);
sort(id+1,id+q+1,[&](ll a, ll b){return l[a]/S==l[b]/S ? r[a]<r[b] : l[a]/S<l[b]/S; });
bit[0].init(1e6), bit[1].init(1e6);
L=1, R=0;
rep(i,1,q)
{
for(;L>l[id[i]];L--)upd(L-1,+1);
for(;R<r[id[i]];R++)upd(R+1,+1);
for(;L<l[id[i]];L++)upd(L,-1);
for(;R>r[id[i]];R--)upd(R,-1);
ans[id[i]]=bit[type[id[i]]].sum(y[id[i]]);
}
rep(i,1,q)
{
printf("%lld\n",ans[i]);
}
return 0;
}