【BZOJ3252】攻略

题解：

首先贪心的会发现我们每次一定会选当前权值和最大的那个

然后在于怎么维护这个最大值

我们发现每个修改实际上是对沿途所有点的子树的修改

所以用线段树维护就可以了。。

另外注意有重复部分，但一定是包含关系所以比较好处理

代码：

#include <bits/stdc++.h>
using namespace std;
#define IL inline
#define ll long long
#define rint register ll
#define rep(i,h,t) for (rint i=h;i<=t;i++)
#define dep(i,t,h) for (rint i=t;i>=h;i--)
#define mid ((h+t)/2)
char ss[1<<24],*A=ss,*B=ss;
char gc()
{
  return A==B&&(B=(A=ss)+fread(ss,1,1<<24,stdin),A==B)?EOF:*A++;
}
template<class T> void read(T &x)
{
  rint f=1,c; while (c=gc(),c<48||c>57) if (c=='-') f=-1; x=c^48;
  while (c=gc(),47<c&&c<58) x=(x<<3)+(x<<1)+(c^48); x*=f; 
}
const ll INF=1e9;
const ll N=3e5;
ll n,m,k,l,v[N],head[N],maxn[N],minn[N],sum[N],fa[N],cnt,ans;
struct re{
  ll a,b;
}a[N];
const ll N2=N*4;
ll data1[N2],data2[N2],data3[N2],lazy1[N2],lazy2[N2],real2[N2];
void arr(ll x,ll y)
{
  a[++l].a=head[x];
  a[l].b=y;
  head[x]=l;
}
IL void down(ll x)
{
  if (!lazy1[x]) return;
  lazy1[x*2]=lazy1[x*2+1]=lazy1[x];
  lazy2[x*2]+=lazy2[x];
  lazy2[x*2+1]+=lazy2[x];
  data3[x*2]+=lazy2[x]; data3[x*2+1]+=lazy2[x];
  data2[x*2]=lazy1[x]; data2[x*2+1]=lazy1[x];
  lazy1[x]=lazy2[x]=0;
}
IL void updata(ll x)
{
  if (data3[x*2]<data3[x*2+1])
  {
    data2[x]=data2[x*2+1];
    data1[x]=data1[x*2+1];
    data3[x]=data3[x*2+1]; 
  } else
  {
    data2[x]=data2[x*2];
    data1[x]=data1[x*2];
    data3[x]=data3[x*2];
  }
}
void build(ll x,ll h,ll t)
{
  if (h==t)
  {
    data1[x]=real2[h]; return;
  }
  build(x*2,h,mid); build(x*2+1,mid+1,t);
}
void change(ll x,ll h,ll t,ll h1,ll t1,ll k1,ll k2)
{
  if (h1<=h&&t<=t1)
  {
    lazy1[x]=k1; lazy2[x]+=k2; data3[x]+=k2; data2[x]=k1;
    return;
  }
  down(x);
  if (h1<=mid) change(x*2,h,mid,h1,t1,k1,k2);
  if (mid<t1)  change(x*2+1,mid+1,t,h1,t1,k1,k2);
  updata(x);
}
void dfs1(ll x)
{
  ll u=head[x];
  if (!u)
  {
    real2[++cnt]=x;
    return;
  }
  while (u)
  {
    ll vv=a[u].b;
    dfs1(vv);
    u=a[u].a;
  }
}
void dfs(ll x,ll y)
{
  ll u=head[x]; sum[x]=y+v[x];
  if (!u)
  {
    minn[x]=maxn[x]=++cnt;
    real2[cnt]=x;
    change(1,1,n,cnt,cnt,0,sum[x]);
    return;
  }
  while (u)
  {
    ll vv=a[u].b;
    dfs(vv,y+v[x]);
    minn[x]=min(minn[x],minn[vv]);
    maxn[x]=max(maxn[x],maxn[vv]);
    u=a[u].a;
  }
}
int main()
{
  freopen("1.in","r",stdin);
  freopen("1.out","w",stdout); 
  read(n); read(k);
  rep(i,1,n) read(v[i]);
  rep(i,1,n) minn[i]=INF;
  rep(i,1,n-1)
  {
    ll x,y;
    read(x); read(y); arr(x,y); fa[y]=x;
  }
  dfs1(1);
  build(1,1,n); cnt=0;
  dfs(1,0);
  rep(i,1,k)
  {
    ll x=data1[1],y=data2[1],z=data3[1];
    ans+=z;
    int kk1=minn[x]-1,kk2=minn[x]; 
    while (x!=y)
    {
      if (kk1>=minn[x]) change(1,1,n,minn[x],kk1,x,-(sum[x]-sum[y]));
      if (kk2<=maxn[x]) change(1,1,n,kk2,maxn[x],x,-(sum[x]-sum[y]));
      kk1=minn[x]-1; kk2=maxn[x]+1;
      x=fa[x];
    }
  }
  cout<<ans<<endl;
  return 0;
}