点分治板子

https://ac.nowcoder.com/acm/contest/11174/E

正好用这道题重新搞一下点分治板子

这道题就是个裸的点分治题

主程序里的话要保留

//#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

//#include <immintrin.h>
//#include <emmintrin.h>
#include <bits/stdc++.h>
using namespace std;
#define rep(i,h,t) for (int i=h;i<=t;i++)
#define dep(i,t,h) for (int i=t;i>=h;i--)
#define ll long long
#define me(x) memset(x,0,sizeof(x))
#define IL inline
#define rint register int
inline ll rd(){
    ll x=0;char c=getchar();bool f=0;
    while(!isdigit(c)){if(c=='-')f=1;c=getchar();}
    while(isdigit(c)){x=(x<<1)+(x<<3)+(c^48);c=getchar();}
    return f?-x:x;
}
char ss[1<<24],*A=ss,*B=ss;
IL char gc()
{
    return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T>void maxa(T &x,T y)
{
    if (y>x) x=y;
}
template<class T>void mina(T &x,T y)
{
    if (y<x) x=y;
}
template<class T>void read(T &x)
{
    int f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=(c^48);
    while(c=gc(),c>47&&c<58) x=x*10+(c^48); x*=f;
}
const int mo=998244353;
ll fsp(int x,int y)
{
    if (y==1) return x;
    ll ans=fsp(x,y/2);
    ans=ans*ans%mo;
    if (y%2==1) ans=ans*x%mo;
    return ans;
}
struct cp {
    ll x,y;
    cp operator +(cp B)
    {
        return (cp){x+B.x,y+B.y};
    }
    cp operator -(cp B)
    {
        return (cp){x-B.x,y-B.y};
    }
    ll operator *(cp B)
    {
        return x*B.y-y*B.x;
    }
    int half() { return y < 0 || (y == 0 && x < 0); }
};
const int N=3e5;
struct re{
    int a,b,c;
}a[N];
int gg[N];
int av[N];
bool vis[N];
ll c[N];
int n,m,rt,son[N],f[N],sum,d[N];
vector<int> pq[N];
ll ans=0;
bool cmp(re x,re y)
{
    return x.a<y.a;
}
#define mid ((h+t)>>1)
struct sgt{
    vector<int> ve; 
    ll sum[N*4];
    int v[N*4];
    void clear()
    {
        for (auto u:ve)
        {
            sum[u]=v[u]=0;
        }
        ve.clear();
    }
    void change(int x,int h,int t,int pos,ll k)
    {
        ve.push_back(x);
        sum[x]=(sum[x]+k)%mo; v[x]++;
        if (h==t) return;
        if (pos<=mid) change(x*2,h,mid,pos,k);
        else change(x*2+1,mid+1,t,pos,k);
    }
    ll q1(int x,int h,int t,int h1,int t1)
    {
        if (h1<=h&&t<=t1) return sum[x];
        ll ans=0;
        if (h1<=mid) ans+=q1(x*2,h,mid,h1,t1);
        if (mid<t1) ans+=q1(x*2+1,mid+1,t,h1,t1);
        ans%=mo;
        return ans; 
    }
    ll q2(int x,int h,int t,int h1,int t1)
    {
        if (h1<=h&&t<=t1) return v[x];
        ll ans=0;
        if (h1<=mid) ans+=q2(x*2,h,mid,h1,t1);
        if (mid<t1) ans+=q2(x*2+1,mid+1,t,h1,t1);
        return ans; 
    }
}S;
void gr(int x,int y)
{
  son[x]=1;f[x]=0;
  for (auto v:pq[x])
    if (vis[v]&&v!=y)
    {
      gr(v,x);
      son[x]+=son[v];
      f[x]=max(f[x],son[v]);
    }
  f[x]=max(f[x],sum-son[x]);
  if (f[x]<f[rt]) rt=x;
}
vector<re> ve,an;
void gd(int x,int fa,int mx,int mn)
{
  ve.push_back((re){mx,mn});
  for (auto v:pq[x])
    if (vis[v]&&v!=fa)
    {
      gd(v,x,max(mx,gg[v]),min(mn,gg[v]));
    }
}
ll gao(vector<re> ve)
{
    S.clear();
    sort(ve.begin(),ve.end(),cmp);
    ll ans=0;
    for (auto v:ve)
    { 
        ans=(ans+S.q1(1,1,n,1,v.b)%mo*c[v.a])%mo;
        if (v.b!=n) 
          ans=(ans+(c[v.a]*c[v.b])%mo*S.q2(1,1,n,v.b+1,n))%mo;
        S.change(1,1,n,v.b,c[v.b]);  
    }
    return ans;
}
void solve(int x,int y)
{
  vis[x]=0;
  an.clear();
  for (auto v:pq[x])
      if (vis[v]&&v!=x)
      {
          gd(v,x,max(gg[v],gg[x]),min(gg[v],gg[x]));
          ans=(ans-gao(ve))%mo;
          an.insert(an.end(),ve.begin(),ve.end());
          ve.clear();
      }
  an.push_back((re){gg[x],gg[x]});
  ans=(ans+gao(an))%mo;
  for (auto v:pq[x])
    if (vis[v])
    {
      rt=0; sum=son[v];
      gr(v,x);
      solve(rt,y+1);
    }
}
int main()
{
    freopen("2.in","r",stdin); 
    freopen("1.out","w",stdout);
   ios::sync_with_stdio(false);
   cin>>n;
   rep(i,1,n) cin>>av[i];
   rep(i,1,n) c[i]=av[i];
   sort(c+1,c+n+1);
   f[0]=1e9;
   int nn=unique(c+1,c+n+1)-c-1;
   rep(i,1,n) gg[i]=lower_bound(c+1,c+nn+1,av[i])-c;
   rep(i,1,n) vis[i]=1;
   rep(i,1,n-1)
   {
        int x,y;
        cin>>x>>y;
        pq[x].push_back(y);
        pq[y].push_back(x);
   }
   sum=n;
   gr(1,0);
   solve(rt,0);
   rep(i,1,n) ans=(ans+1ll*av[i]*av[i])%mo;
   cout<<(ans+mo)%mo<<endl;
   return 0;
}
View Code

 

posted @ 2021-06-11 22:40  尹吴潇  阅读(34)  评论(0编辑  收藏  举报