BZOJ2809: [Apio2012]dispatching

Posted 2020-08-22

传送门

主席树经典题。

首先把树搞出来，然后搞出来DFS序。然后离散化点权，在DFS序上建立主席树。

对于每个点对应的区间，查找对应的区间最大的点数即可。

//BZOJ2809
//by Cydiater
//2016.12.6
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
#include <ctime>
#include <cmath>
#include <iomanip>
#include <bitset>
#include <set>
using namespace std;
#define ll long long
#define up(i,j,n)		for(int i=j;i<=n;i++)
#define down(i,j,n)		for(int i=j;i>=n;i--)
#define cmax(a,b)		a=max(a,b)
#define cmin(a,b)		a=min(a,b)
#define Auto(i,node)		for(int i=LINK[node];i;i=e[i].next)
#define FILE "dispatching"
const int MAXN=1e5+5;
const ll oo=1LL<<55;
inline int read(){
	char ch=getchar();int x=0,f=1;
	while(ch>‘9‘||ch<‘0‘){if(ch==‘-‘)f=-1;ch=getchar();}
	while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
	return x*f;
}
int N,root[MAXN],cnt=0,siz[MAXN],dfn[MAXN],dfs_clock=0;
ll M,va[MAXN],vb[MAXN],fsort[MAXN],rnum,val[MAXN],ans=0;
struct Graph{
	int LINK[MAXN],len;
	struct edge{
		int y,next;
	}e[MAXN<<1];
	inline void insert(int x,int y){e[++len].next=LINK[x];LINK[x]=len;e[len].y=y;}
	inline void Insert(int x,int y){insert(x,y);insert(y,x);}
	void make_graph(){
		up(i,1,N){
			int fa=read();va[i]=read();vb[i]=read();
			fsort[i]=va[i];
			if(!fa)continue;
			Insert(i,fa);
		}
	}
	void dfs(int node,int father){
		siz[node]=1;dfn[++dfs_clock]=node;
		Auto(i,node)if(e[i].y!=father){
			dfs(e[i].y,node);
			siz[node]+=siz[e[i].y];
		}
	}
}G;
struct Chair_man_Tree{
	ll cost,sum;
	int son[2];
}t[MAXN<<5];
namespace solution{
	void init(){
		N=read();M=read();
		G.make_graph();
		G.dfs(1,0);
	}
	int NewNode(ll cost,int sum,int son0,int son1){
		t[++cnt].cost=cost;t[cnt].sum=sum;
		t[cnt].son[0]=son0;t[cnt].son[1]=son1;
		return cnt;
	}
	void insert(int leftt,int rightt,int &Root,int last,int pos){
		Root=NewNode(t[last].cost+fsort[pos],t[last].sum+1,t[last].son[0],t[last].son[1]);
		int mid=(leftt+rightt)>>1;
		if(leftt==rightt)	return;
		if(pos<=mid)		insert(leftt,mid,t[Root].son[0],t[last].son[0],pos);
		else			insert(mid+1,rightt,t[Root].son[1],t[last].son[1],pos);
	}
	ll Get(int S,int T,int leftt,int rightt,ll LIM){
		if(leftt==rightt)		return min(LIM/fsort[leftt],t[T].sum-t[S].sum);
		int mid=(leftt+rightt)>>1;
		ll cost=t[t[T].son[0]].cost-t[t[S].son[0]].cost;
		if(cost<=LIM)	return t[t[T].son[0]].sum-t[t[S].son[0]].sum+Get(t[S].son[1],t[T].son[1],mid+1,rightt,LIM-cost);
		else		return Get(t[S].son[0],t[T].son[0],leftt,mid,LIM);
	}
	void slove(){
		sort(fsort+1,fsort+N+1);
		rnum=unique(fsort+1,fsort+N+1)-(fsort+1);
		up(i,1,N)val[i]=lower_bound(fsort+1,fsort+rnum+1,va[dfn[i]])-fsort;
		up(i,1,N)insert(1,rnum,root[i],root[i-1],val[i]);
		up(i,1,N){
			int L=i,R=i+siz[dfn[i]]-1;
			cmax(ans,vb[dfn[i]]*Get(root[L-1],root[R],1,rnum,M));
		}	
	}
	void output(){
		cout<<ans<<endl;
	}
}
int main(){
	//freopen("input.in","r",stdin);
	//freopen(FILE".in","r",stdin);
	//freopen(FILE".out","w",stdout);
	using namespace solution;
	init();
	slove();
	output();
	return 0;
}