BZOJ 2402 陶陶的难题II (树链剖分、线段树、凸包、分数规划)

毒瘤，毒瘤，毒瘤……

(30000)这个数据范围，看上去就是要搞事的啊。。。

题目链接: https://www.lydsy.com/JudgeOnline/problem.php?id=2402

题解:

首先化式子: 假设二分的答案为(mid)则(frac{y_i+q_j}{x_i+p_j}ge mid, y_i+q_jge mid(x_i+p_j), y_i-mid imes x_i+q_j-mid imes p_jge 0)

那么我们实际上就是要找一个(i)使得(-x_i imes mid+y_i)最大，同理找一个(j)使得(-p_j imes mid+q_j)最大，这两个任务完全一样，现在就讨论前者

为了方便(强迫症)让我们把它变成(-a_ix+b_i)的形式

仿照斜率优化的思路，如果(a_i<a_j)，那么(i)不劣于(j)当且仅当(-a_ix+b_ige -a_jx+b_j, xge frac{b_j-b_i}{a_j-a_i}).

那么假设有并排着的三个点(i,j,k,a_i<a_j<a_k), 则(i)不劣于(j)当且仅当(xge frac{b_j-b_i}{a_j-a_i}), (k)不劣于(j)当且仅当(xle frac{b_k-b_j}{a_k-a_j}), 如果(frac{b_j-b_i}{a_j-a_i}le frac{b_k-b_j}{a_k-a_j}), 那么对于任何(x)上面两个条件至少有一个成立，那么(j)就是无用的。也就是说有用的点一定满足斜率递减，在上凸壳的左半边上。

这题没有修改操作，那么我们可以用线段树求出每个子区间内的凸壳，然后对于每个询问，先分数规划二分答案，然后求树链剖分划分成的每一段区间的最大值，这个可以通过在线段树每个子区间的凸壳上二分得到。

时间复杂度(O(nlog^4n))

(这可能是我最近写的几道题唯一一道跑进BZOJ前一半的？23333)

代码

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<vector>
using namespace std;

const int N = 3e4;
const double EPS = 1e-8;
const double INF = 1e6;
int dcmp(double x) {return x<-EPS ? -1 : (x>EPS ? 1 : 0);}
struct Point
{
	double x,y;
	Point() {}
	Point(double _x,double _y) {x = _x,y = _y;}
	bool operator <(const Point &arg) const {return dcmp(x-arg.x)<0 || (dcmp(x-arg.x)==0 && dcmp(y-arg.y)<0);}
};
struct Edge
{
	int v,w,nxt;
} e[(N<<1)+3];
int fe[N+3];
int fa[N+3];
int dfn[N+3];
int idfn[N+3];
int sz[N+3];
int tpn[N+3];
int dep[N+3];
int hvs[N+3];
double a1[N+3],b1[N+3],a2[N+3],b2[N+3];
int n,q,en,cnt;

struct SegmentTree
{
	vector<Point> sgt[(N<<2)+3];
	void CHpush(int id,Point x)
	{
		int j = sgt[id].size()-1;
		while(j>0 && dcmp((sgt[id][j].y-sgt[id][j-1].y)*(x.x-sgt[id][j].x)-(sgt[id][j].x-sgt[id][j-1].x)*(x.y-sgt[id][j].y))<=0)
		{
			sgt[id].pop_back();
			j--;
		}
		sgt[id].push_back(x);
	}
	void build(int rtn,int le,int ri,double a[],double b[])
	{
		if(le==ri)
		{
			sgt[rtn].push_back(Point(a[idfn[le]],b[idfn[le]]));
			return;
		}
		int mid = (le+ri)>>1;
		build(rtn<<1,le,mid,a,b);
		build(rtn<<1|1,mid+1,ri,a,b);
		int i = 0,j = 0;
		while(i<sgt[rtn<<1].size() && j<sgt[rtn<<1|1].size())
		{
			if(sgt[rtn<<1][i]<sgt[rtn<<1|1][j]) {CHpush(rtn,sgt[rtn<<1][i]); i++;}
			else {CHpush(rtn,sgt[rtn<<1|1][j]); j++;}
		}
		while(i<sgt[rtn<<1].size()) {CHpush(rtn,sgt[rtn<<1][i]); i++;}
		while(j<sgt[rtn<<1|1].size()) {CHpush(rtn,sgt[rtn<<1|1][j]); j++;}
	}
	double calc(int rtn,double x)
	{
		int left = 0,right = sgt[rtn].size()-1;
		while(left<right)
		{
			int mid = (left+right+1)>>1;
			if(mid==0) {break;}
			if(dcmp(x*(sgt[rtn][mid].x-sgt[rtn][mid-1].x)-(sgt[rtn][mid].y-sgt[rtn][mid-1].y))<=0) {left = mid;}
			else {right = mid-1;}
		}
		double ret = -sgt[rtn][left].x*x+sgt[rtn][left].y;
		return ret;
	}
	double query(int rtn,int le,int ri,int lb,int rb,double x)
	{
		if(le>=lb && ri<=rb) {return calc(rtn,x);}
		int mid = (le+ri)>>1; double ret = -INF;
		if(lb<=mid) {ret = max(ret,query(rtn<<1,le,mid,lb,rb,x));}
		if(rb>mid) {ret = max(ret,query(rtn<<1|1,mid+1,ri,lb,rb,x));}
		return ret;
	}
} sgt1,sgt2;

void addedge(int u,int v)
{
	en++; e[en].v = v;
	e[en].nxt = fe[u]; fe[u] = en;
}

void dfs1(int u)
{
	sz[u] = 1; hvs[u] = 0;
	for(int i=fe[u]; i; i=e[i].nxt)
	{
		int v = e[i].v;
		if(v==fa[u]) continue;
		fa[v] = u;
		dep[v] = dep[u]+1;
		dfs1(v);
		sz[u] += sz[v];
		if(sz[v]>sz[hvs[u]]) {hvs[u] = v;}
	}
}

void dfs2(int u)
{
	cnt++; dfn[u] = cnt; idfn[cnt] = u;
	if(hvs[u]) {tpn[hvs[u]] = tpn[u]; dfs2(hvs[u]);}
	for(int i=fe[u]; i; i=e[i].nxt)
	{
		int v = e[i].v;
		if(v==hvs[u] || v==fa[u]) continue;
		tpn[v] = v;
		dfs2(v);
	}
}

bool query(int u,int v,double x)
{
	double ret1 = -INF,ret2 = -INF;
	while(tpn[u]!=tpn[v])
	{
		if(dep[fa[tpn[u]]]>dep[fa[tpn[v]]])
		{
			double tmp = sgt1.query(1,1,n,dfn[tpn[u]],dfn[u],x);
			ret1 = max(ret1,tmp);
			tmp = sgt2.query(1,1,n,dfn[tpn[u]],dfn[u],x);
			ret2 = max(ret2,tmp);
			u = fa[tpn[u]];
		}
		else
		{
			double tmp = sgt1.query(1,1,n,dfn[tpn[v]],dfn[v],x);
			ret1 = max(ret1,tmp);
			tmp = sgt2.query(1,1,n,dfn[tpn[v]],dfn[v],x);
			ret2 = max(ret2,tmp);
			v = fa[tpn[v]];
		}
	}
	if(dep[u]>dep[v]) swap(u,v);
	double tmp = sgt1.query(1,1,n,dfn[u],dfn[v],x);
	ret1 = max(ret1,tmp);
	tmp = sgt2.query(1,1,n,dfn[u],dfn[v],x);
	ret2 = max(ret2,tmp);
	if(dcmp(ret1+ret2)>0) return true;
	return false;
}

int main()
{
	scanf("%d",&n);
	for(int i=1; i<=n; i++) scanf("%lf",&a1[i]);
	for(int i=1; i<=n; i++) scanf("%lf",&b1[i]);
	for(int i=1; i<=n; i++) scanf("%lf",&a2[i]);
	for(int i=1; i<=n; i++) scanf("%lf",&b2[i]);
	for(int i=1; i<n; i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		addedge(x,y); addedge(y,x);
	}
	dep[1] = 1; dfs1(1);
	tpn[1] = 1; dfs2(1);
	sgt1.build(1,1,n,a1,b1);
	sgt2.build(1,1,n,a2,b2);
	scanf("%d",&q);
	for(int i=1; i<=q; i++)
	{
		int x,y; scanf("%d%d",&x,&y);
		double left = 0.0,right = 1e5;
		while(right-left>5e-5)
		{
			double mid = (left+right)*0.5;
			bool ans = query(x,y,mid);
			if(ans) {left = mid;}
			else {right = mid;}
		}
		printf("%lf
",left);
	}
	return 0;
}