题解 \(by\;zj\varphi\)
答案可以转化为,每个点的子树中小于子树的根的数量的和。
可以先用权值树状数组求出来 1 的答案,然后再用换根 dp 解决。
\(dp_v=dp_x+low_{1,v}-low_{v,v}-low_{v,x}\)
其中 \(low_{x,v}\) 表示在以 x 为根的子树中,有多少小于 \(w_v\) 的值。
复杂度 \(\mathcal{O\rm{(nlogn)}}\)。
Code#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
char buf[1<<21],*p1=buf,*p2=buf;
#define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
#define debug1(x) std::cerr << #x"=" << x << ' '
#define debug2(x) std::cerr << #x"=" << x << std::endl
#define Debug(x) assert(x)
struct nanfeng_stream{
template<typename T>inline nanfeng_stream &operator>>(T &x) {
bool f=false;x=0;char ch=gc();
while(!isdigit(ch)) f|=ch=='-',ch=gc();
while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
return x=f?-x:x,*this;
}
}cin;
}
using IO::cin;
namespace nanfeng{
#define FI FILE *IN
#define FO FILE *OUT
template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
using ll=long long;
static const int N=1e6+7;
struct edge{int v,nxt,w;}e[N<<1];
int first[N],t=1,st[N],w[N],siz[N],n,q;
ll ans[N],as;
auto add=[](int u,int v) {
e[t]={v,first[u],0},first[u]=t++;
e[t]={u,first[v],0},first[v]=t++;
};
struct BIT{
#define lowbit(x) ((x)&-(x))
int c[N];
func(void(void)) init=[&]() {memset(c,0,sizeof(c));};
func(void(int,int)) update=[&](int x,int k) {for (;x<=n;x+=lowbit(x)) c[x]+=k;};
func(int(int)) query=[&](int x) {
// Debug(x>=0);
int res=0;
for (;x;x-=lowbit(x)) res+=c[x];
return res;
};
}B;
func(void(int,int)) dfs=[](int x,int fa) {
B.update(w[x],1);
int lft=B.query(w[x]-1),kl=lft;
for (ri i(first[x]),v;i;i=e[i].nxt) {
if ((v=e[i].v)==fa) continue;
dfs(v,x);
int nw=B.query(w[x]-1);
e[i].w=nw-kl;
kl=nw;
}
siz[x]=B.query(w[x]-1)-lft;
ans[1]+=siz[x];
};
func(void(int,int)) solve=[](int x,int fa) {
for (ri i(first[x]),v;i;i=e[i].nxt) {
if ((v=e[i].v)==fa) continue;
ans[v]=ans[x]-e[i].w+B.query(w[v]-1)-siz[v];
solve(v,x);
}
};
inline int main() {
FI=freopen("tears.in","r",stdin);
FO=freopen("tears.out","w",stdout);
cin >> n >> q;
for (ri i(1);i<=n;pd(i)) cin >> w[i],st[i]=w[i];
std::sort(st+1,st+n+1);
int k=std::unique(st+1,st+n+1)-st;
for (ri i(1);i<=n;pd(i)) w[i]=std::lower_bound(st+1,st+k,w[i])-st;
for (ri i(1),u,v;i<n;pd(i)) cin >> u >> v,add(u,v);
dfs(1,0);
solve(1,0);
for (ri i(1),u;i<=q;pd(i)) cin >> u,printf("%lld\n",ans[u]);
return 0;
}
}
int main() {return nanfeng::main();}