[堆 折线] UOJ#205 & BZOJ4585 【APIO2016】Fireworks

大概是一个子树中的 fx 是一个单峰函数，转折点用可并堆维护

具体看这里

#include <cstdio>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int N=1200010;int n,m,cnt,fa[N],G[N];struct edge{  int t,w,nx;}E[N];inline char nc(){  static char buf[100000],*p1=buf,*p2=buf;  return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline void read(int &x){  char c=nc(); x=0;  for(;c>'9'||c<'0';c=nc());for(;c>='0'&&c<='9';x=x*10+c-'0',c=nc());}inline void addedge(int x,int y,int w){  E[++cnt].t=y; E[cnt].nx=G[x]; G[x]=cnt; E[cnt].w=w;}struct node{  node *l,*r;  ll v;}p[N],*rt[N],*t;inline int ran(){  static int x=31253125; x+=(x<<4)+1; return x&65536;}inline node *New(ll x){  t->v=x; t->l=t->r=0;  return t++;}node *Merge(node *x,node *y){  if(!x || !y) return !x?y:x;  if(x->v<y->v) swap(x,y);  ran()?x->l=Merge(x->l,y):x->r=Merge(x->r,y);  return x;}inline void Pop(int v){  rt[v]=Merge(rt[v]->l,rt[v]->r);}ll ans;int main(){  freopen("1.in","r",stdin);  freopen("1.out","w",stdout);  read(n); read(m); t=p;  for(int i=2,p,w;i<=n+m;i++)    read(p),read(w),addedge(p,i,w),ans+=w;  for(int i=n+1;i<=n+m;i++)    rt[i]=Merge(rt[i],Merge(New(0),New(0)));  for(int x=n;x;x--){    int d=0;    for(int i=G[x];i;i=E[i].nx){      int v=E[i].t; ll L,R;      R=rt[v]->v; Pop(v);      L=rt[v]->v,Pop(v);      rt[v]=Merge(rt[v],New(L+E[i].w));      rt[v]=Merge(rt[v],New(R+E[i].w));      rt[x]=Merge(rt[x],rt[v]);      d++;    }    for(int i=1;i<d;i++) Pop(x);  }  Pop(1);  while(rt[1]) ans-=rt[1]->v,Pop(1);  cout<<ans<<endl;  return 0;}