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【BZOJ】【P3672】【Noi2014】【购票】【题解】【线段树+凸包+链剖+三分】

2017年04月19日 ⁄ 综合 ⁄ 共 3136字 ⁄ 字号 评论关闭

传送门:http://www.lydsy.com/JudgeOnline/problem.php?id=3672

我写的是nlog^3n的链剖凸包,题解有很多,扔一发代码

Code:

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<cctype>
typedef long long LL;
const int maxn=2e5+5;
using namespace std;
int DEB;
LL getint(){
	LL res=0;char c=getchar();
	while(!isdigit(c))c=getchar();
	while(isdigit(c))res=res*10+c-'0',c=getchar();
	return res;
}
struct point{
	LL x,y;
	point(LL _x=0,LL _y=0):x(_x),y(_y){}
	double operator*(point oth)const{return (double)x*oth.y-(double)y*oth.x;}
	LL operator^(point oth)const{return (LL)x*oth.x+(LL)y*oth.y;}
	point operator-(point oth)const{return point(x-oth.x,y-oth.y);}
	point operator+(point oth)const{return point(x+oth.x,y+oth.y);}
	bool operator<(point oth)const{return x<oth.x||(x==oth.x&&y<oth.y);}
	void print(){printf("%d %d\n",int(x),int(y));}
}tmp[maxn];
int tmpsize,n,T,siz[maxn],son[maxn],dep[maxn],top[maxn],w[maxn],rw[maxn],fa[maxn],z,dfn[maxn];
LL d[maxn],p[maxn],q[maxn],l[maxn],dp[maxn];
vector<int>G[maxn];
void dfs(int u){
	siz[u]=1;dfn[++dfn[0]]=u;
	for(int i=0;i<G[u].size();i++){
		int v=G[u][i];
		if(v!=fa[u]){
			fa[v]=u;dep[v]=dep[u]+1;d[v]+=d[u];
			dfs(v);
			siz[u]+=siz[v];
			if(siz[son[u]]<siz[v])son[u]=v;
		}
	}
}
void buiLL(int u,int tp){
	w[u]=++z;rw[z]=u;top[u]=tp;
	if(son[u])buiLL(son[u],tp);
	for(int i=0;i<G[u].size();i++){
		int v=G[u][i];
		if(v!=fa[u]&&v!=son[u])buiLL(v,v);
	}
}
struct CH{
	point *ch;
	int size;
	void init(int l,int r){
		tmpsize=size=0;
		for(int i=l;i<=r;i++)tmp[++tmpsize]=point(d[rw[i]],dp[rw[i]]);
		sort(tmp+1,tmp+1+tmpsize);
		ch=new point[tmpsize+5];
		for(int i=1;i<=tmpsize;i++){
			while(size>1&&(tmp[i]-ch[size])*(ch[size]-ch[size-1])>=-1e-20)size--;
			ch[++size]=tmp[i];			
		}
	}
	LL qmin(point p){
		int l=1,r=size;
		LL ans=min((ch[l]^p),(ch[r]^p));
		while(r-l>2){
			int mid1=l+(r-l)/3;
			int mid2=r-(r-l)/3;
			if((ch[mid1]^p)<=(ch[mid2]^p))
				r=mid2;
			else l=mid1;
		}		
		for(int i=l;i<=r;i++)ans=min(ans,(LL)(ch[i]^p));
		return ans;
	}
};
int findrt(int u){
	LL lim=d[u]-l[u];
	for(;u&&d[top[u]]>=lim;){
		u=top[u];
		if(d[fa[u]]>=lim)u=fa[u];else break;
	}
	if(!u)return 1;
	int l=w[u],r=w[top[u]];
	if(l>r)swap(l,r);
	while(l<r){
		int mid=(l+r)>>1;
		if(d[rw[mid]]<lim)
			l=mid+1;
		else r=mid;
	}return rw[l];
}
bool bud[maxn<<2];
CH t[maxn<<2];
LL Qmin(int i,int l,int r,int l0,int r0,point p){
	if(l0>r0)swap(l0,r0);
	if(l0<=l&&r0>=r){
		if(bud[i])return t[i].qmin(p);
		bud[i]=1;t[i].init(l,r);
		return t[i].qmin(p);
	}int mid=(l+r)>>1;LL ans=1LL<<61;
	if(l0<=mid)ans=min(ans,Qmin(i<<1,l,mid,l0,r0,p));
	if(r0>mid)ans=min(ans,Qmin(i<<1|1,mid+1,r,l0,r0,p));
	return ans;
}
int main(){
	n=getint();T=getint();
	for(int i=2;i<=n;i++){
		fa[i]=getint(),d[i]=getint(),p[i]=getint(),q[i]=getint(),l[i]=getint();
		G[i].push_back(fa[i]);G[fa[i]].push_back(i);
	}if(T==0||T==2){
		queue<int>q;
		q.push(1);
		while(!q.empty()){
			int u=q.front();q.pop();
			dfn[++dfn[0]]=u;w[u]=++z;rw[z]=u;top[u]=1;
			for(int i=0;i<G[u].size();i++){
				int v=G[u][i];
				if(v!=fa[u]){
					q.push(v);
					d[v]+=d[u];fa[v]=u;
					dep[v]=dep[u]+1;
					son[u]=v;					
				}			
			}
		}
	}else dfs(1),buiLL(1,1);	
	for(int i=2;i<=dfn[0];i++){
		DEB=dfn[i];
		dp[dfn[i]]=1LL<<61;int u=fa[dfn[i]],v=findrt(dfn[i]);
		for(;top[u]!=top[v];u=fa[top[u]])
			dp[dfn[i]]=min(dp[dfn[i]],Qmin(1,1,n,w[u],w[top[u]],point(-p[dfn[i]],1)));
		dp[dfn[i]]=min(dp[dfn[i]],Qmin(1,1,n,w[u],w[v],point(-p[dfn[i]],1)));
		dp[dfn[i]]+=p[dfn[i]]*d[dfn[i]]+q[dfn[i]];
	}
	for(int i=2;i<=n;i++)printf("%lld\n",dp[i]);
	return 0;
}

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