【bzoj3672】[Noi2014]购票 斜率优化+树链剖分+线段树+凸包+三分

f[i]表示从根到点i的最少票价
f[i]=min{f[j]+(dep[i]-dep[j])*p[i]+q[i] } (dep[i]-dep[j]<=li)
=f[j]-dep[j]*p[i]+dep[i]*p[i]+q[i]
f[j]=dep[j]*p[i]+f[i]-dep[i]*p[i]-q[i]
f[i]-dep[i]*p[i]-q[i]表示过点(dep[j],f[j])的斜率为p[i]的直线在y轴上的截距
因为p[i]>=0，所以答案一定在下凸壳上
pre[i]表示i最多能延伸到的祖先，这个可以二分什么的乱搞出来
求f[i]就是在fa[i]到pre[i]之间形成的凸壳上三分

树链剖分+线段树维护凸壳
线段树的每个节点暴力建出凸壳，复杂度O(nlog^2n)

每次查询按照剖分查就可以了，复杂度O(log^3n)

注意，本题算叉积的时候会爆longlong，要转成double

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<vector>
#define maxn 200010 
#define inf 1e18

using namespace std;

struct yts
{
	long long x,y;
}p[maxn],tmp[maxn];

bool operator<(yts x,yts y)
{
	return x.x<y.x || (x.x==y.x && x.y<y.y);
}

double operator*(yts x,yts y)
{
	double X1=x.x,X2=y.x,Y1=x.y,Y2=y.y;
	return X1*Y2-X2*Y1;
}

yts operator-(yts x,yts y)
{
	yts ans;
	ans.x=x.x-y.x;ans.y=x.y-y.y;
	return ans;
}

struct yts1
{
	int l,r;
	int size;
	bool flag;
	vector<yts> v;
}t[4*maxn];

int st[maxn],to[maxn],next[maxn],fa[maxn],pre[maxn],g[20][maxn];
long long len[maxn],dep[maxn],f[maxn];
int seq[maxn],head[maxn],e[maxn],rank[maxn],size[maxn],dd[maxn],d[maxn];
long long P[maxn],Q[maxn],L[maxn];
bool vis[maxn];
int n,m,num,tot,T;

void addedge(int x,int y,long long z)
{
	num++;to[num]=y;len[num]=z;next[num]=st[x];st[x]=num;
}

void dfs1(int x)
{
	e[++tot]=x;dd[x]=0;size[x]=1;
	for (int p=st[x];p;p=next[p])
	{
		dep[to[p]]=dep[x]+len[p];d[to[p]]=d[x]+1;
		dfs1(to[p]);
		size[x]+=size[to[p]];
		if (size[to[p]]>size[dd[x]]) dd[x]=to[p];
	}
}

int cal(int x)
{
	int ans=x;
	for (int k=19;k>=0;k--)
	  if (g[k][ans] && dep[x]-dep[g[k][ans]]<=L[x]) ans=g[k][ans];
	return ans;
}

void build(int i,int l,int r)
{
	t[i].l=l;t[i].r=r;t[i].flag=0;
	yts x;x.x=0;x.y=0;
	for (int j=t[i].l;j<=t[i].r+1;j++) t[i].v.push_back(x);
	if (l==r) return;
	int mid=(l+r)/2;
	build(i*2,l,mid);build(i*2+1,mid+1,r);
}

void build(int i)
{
	int size=0;
	for (int j=t[i].l;j<=t[i].r;j++) tmp[++size]=p[j];
	sort(tmp+1,tmp+size+1);
	t[i].size=0;
	for (int j=1;j<=size;j++)
	{
		while (t[i].size>1 && (tmp[j]-t[i].v[t[i].size])*(t[i].v[t[i].size]-t[i].v[t[i].size-1])>=0) t[i].size--;
		t[i].v[++t[i].size]=tmp[j];
	}
}

long long calc(int i,int j,int id)
{
	return t[i].v[j].y+(dep[id]-t[i].v[j].x)*P[id]+Q[id];
}

long long query(int i,int id)
{
	long long ans=inf;
	int l=1,r=t[i].size;
	while (r-l>=3)
	{
		int mid=l+(r-l)/3,midmid=r-(r-l)/3;
		if (calc(i,mid,id)<calc(i,midmid,id)) r=midmid; else l=mid;
	}
	for (int j=l;j<=r;j++) ans=min(ans,calc(i,j,id));
	return ans;
}

long long query(int i,int l,int r,int id)
{
	if (l<=t[i].l && t[i].r<=r)
	{
		if (!t[i].flag) build(i),t[i].flag=1;
		return query(i,id);
	}
	int mid=(t[i].l+t[i].r)/2;
	long long ans=inf;
	if (l<=mid) ans=min(ans,query(i*2,l,r,id));
	if (mid<r) ans=min(ans,query(i*2+1,l,r,id));
	return ans;
}

long long query(int x,int y,int id)
{
	long long ans=inf;
	while (d[head[x]]>d[y])
	{
		ans=min(ans,query(1,rank[head[x]],rank[x],id));x=fa[head[x]];
	}
	ans=min(ans,query(1,rank[y],rank[x],id));
	return ans;
}

int main()
{
	scanf("%d%d",&n,&T);
	for (int i=2;i<=n;i++)
	{
		int x;
		scanf("%d%lld%lld%lld%lld",&fa[i],&x,&P[i],&Q[i],&L[i]);
		addedge(fa[i],i,x);
	}
	dfs1(1);
	int qwer=0;
	for (int i=1;i<=n;i++)
	  if (!vis[e[i]])
	  {
	  	int k=e[i];
	  	while (k)
	  	{
	  		seq[++qwer]=k;head[k]=e[i];vis[k]=1;k=dd[k];
	  	}
	  }
	for (int i=1;i<=n;i++) rank[seq[i]]=i,g[0][i]=fa[i];
	for (int j=1;j<=18;j++)
	  for (int i=1;i<=n;i++)
	    g[j][i]=g[j-1][g[j-1][i]];
	for (int i=1;i<=n;i++) pre[i]=cal(i);
	build(1,1,n);
	f[1]=0;dep[1]=0;p[rank[1]].x=0;p[rank[1]].y=0;
	for (int i=2;i<=n;i++)
	{
		f[i]=query(fa[i],pre[i],i);
		p[rank[i]].x=dep[i];p[rank[i]].y=f[i];
		printf("%lld\n",f[i]);
	}
	return 0;
}