题目大意：给定一棵树，选定一棵子树中的一些点，薪水和不能超过m，求点的数量*子树根节点的领导能力的最大值
考虑对于每一个节点，我们维护一种数据结构，在当中贪心寻找薪金小的雇佣。
每一个节点暴力重建一定不行。我们考虑可并数据结构。每一个节点将子节点的信息直接合并就可以
能够用启示式合并的Treap。也能够用可并堆
今天特意去学了这玩应0.0 先写了左偏树 然后又写了下随机堆…… 后者速度上更快一些 只是建议从左偏树開始学起
总之平衡树常数各种大就是了0.0
Treap+启示式合并
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 100100
using namespace std;
typedef long long ll;
struct abcd{
abcd *ls,*rs;
int key;
int cnt,siz;
ll num,sum;
abcd (ll x,int y);
void Maintain();
}*null=new abcd(0,0),*tree[M];
struct edge{
int to,next;
}table[M];
int head[M],tot;
int n,root;
ll m,ans,leadership[M];
void Add(int x,int y)
{
table[++tot].to=y;
table[tot].next=head[x];
head[x]=tot;
}
abcd :: abcd(ll x,int y)
{
ls=rs=null;
sum=x*y;
num=x;
cnt=siz=y;
key=y?rand():0;
}
void abcd :: Maintain()
{
siz=ls->siz+rs->siz+cnt;
sum=ls->sum+rs->sum+num*cnt;
}
void Zig(abcd *&x)
{
abcd *y=x->ls;
x->ls=y->rs;
y->rs=x;
x=y;
x->rs->Maintain();
}
void Zag(abcd *&x)
{
abcd *y=x->rs;
x->rs=y->ls;
y->ls=x;
x=y;
x->ls->Maintain();
}
void Insert(abcd *&x,ll y,int z)
{
if(x==null)
{
x=new abcd(y,z);
return ;
}
if(y==x->num)
x->cnt+=z;
else if(y<x->num)
{
Insert(x->ls,y,z);
if(x->ls->key>x->key)
Zig(x);
}
else
{
Insert(x->rs,y,z);
if(x->rs->key>x->key)
Zag(x);
}
x->Maintain();
}
int Query(abcd *x,ll y)
{
if(x==null)
return 0;
ll temp=x->ls->sum;int re=0;
if(y<=temp) return Query(x->ls,y);
re+=x->ls->siz;y-=temp;
if(y<=x->num*x->cnt)
return re+y/x->num;
re+=x->cnt;y-=x->num*x->cnt;
return re+Query(x->rs,y);
}
void Decomposition(abcd *&x,int y)
{
if(x==null)
return ;
Decomposition(x->ls,y);
Decomposition(x->rs,y);
Insert(tree[y],x->num,x->cnt);
delete x;
x=null;
}
void Tree_DP(int x)
{
int i;
for(i=head[x];i;i=table[i].next)
{
Tree_DP(table[i].to);
if(tree[x]->siz<tree[table[i].to]->siz)
swap(tree[x],tree[table[i].to]);
Decomposition(tree[table[i].to],x);
}
ans=max(ans,leadership[x]*Query(tree[x],m));
}
int main()
{ //freopen("2809.in","r",stdin);
//freopen("2809.out","w",stdout); int i,fa;
ll x;
cin>>n>>m;
for(i=1;i<=n;i++)
{
scanf("%d%lld%lld",&fa,&x,&leadership[i]);
if(!fa) root=i;
else Add(fa,i);
tree[i]=new abcd(x,1);
}
Tree_DP(root);
cout<<ans<<endl;
}
//lld!!
左偏树
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 100100
using namespace std;
struct abcd{
abcd *ls,*rs;
int num,h;
abcd(int x);
}*null=new abcd(0),*tree[M];
struct edge{
int to,next;
}table[M];
int head[M],tot;
int n,m,root,leadership[M],sum[M],size[M];
long long ans;
void Add(int x,int y)
{
table[++tot].to=y;
table[tot].next=head[x];
head[x]=tot;
}
abcd :: abcd(int x)
{
ls=rs=null;
num=x;
if(x) h=0;
else h=-1;
}
abcd* Merge(abcd *x,abcd *y)
{
if(x==null) return y;
if(y==null) return x;
if(x->num<y->num)
swap(x,y);
x->rs=Merge(x->rs,y);
if(x->ls->h<x->rs->h)
swap(x->ls,x->rs);
x->h=x->rs->h+1;
return x;
}
void Tree_DP(int x)
{
int i;
for(i=head[x];i;i=table[i].next)
{
Tree_DP(table[i].to);
tree[x]=Merge(tree[x],tree[table[i].to]);
sum[x]+=sum[table[i].to];
size[x]+=size[table[i].to];
while(sum[x]>m)
{
sum[x]-=tree[x]->num;
--size[x];
tree[x]=Merge(tree[x]->ls,tree[x]->rs);
}
}
ans=max(ans,(long long)size[x]*leadership[x]);
}
int main()
{
int i,fa,x;
cin>>n>>m;
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&fa,&x,&leadership[i]);
if(!fa) root=i;
else Add(fa,i);
tree[i]=new abcd(x);
sum[i]=x;
size[i]=1;
}
Tree_DP(root);
cout<<ans<<endl;
}
随机堆
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define M 100100
using namespace std;
struct abcd{
abcd *ls,*rs;
int num;
abcd(int x);
}*null=new abcd(0),*tree[M];
struct edge{
int to,next;
}table[M];
bool son;
int head[M],tot;
int n,m,root,leadership[M],sum[M],size[M];
long long ans;
void Add(int x,int y)
{
table[++tot].to=y;
table[tot].next=head[x];
head[x]=tot;
}
abcd :: abcd(int x)
{
ls=rs=null;
num=x;
}
abcd* Merge(abcd *x,abcd *y)
{
if(x==null) return y;
if(y==null) return x;
if(x->num<y->num)
swap(x,y);
if(son^=1)
x->rs=Merge(x->rs,y);
else
x->ls=Merge(x->ls,y);
return x;
}
void Tree_DP(int x)
{
int i;
for(i=head[x];i;i=table[i].next)
{
Tree_DP(table[i].to);
tree[x]=Merge(tree[x],tree[table[i].to]);
sum[x]+=sum[table[i].to];
size[x]+=size[table[i].to];
while(sum[x]>m)
{
sum[x]-=tree[x]->num;
--size[x];
tree[x]=Merge(tree[x]->ls,tree[x]->rs);
}
}
ans=max(ans,(long long)size[x]*leadership[x]);
}
int main()
{
int i,fa,x;
cin>>n>>m;
for(i=1;i<=n;i++)
{
scanf("%d%d%d",&fa,&x,&leadership[i]);
if(!fa) root=i;
else Add(fa,i);
tree[i]=new abcd(x);
sum[i]=x;
size[i]=1;
}
Tree_DP(root);
cout<<ans<<endl;
}