NOIP 模拟 $83; m 时代的眼泪$

题解 (by;zjvarphi)

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 m{(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
",ans[u]);
        return 0;
    }
}
int main() {return nanfeng::main();}