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\(\frac{y_i+q_j}{x_i+p_j}\ge mid, y_i+q_j\ge mid(x_i+p_j), y_i-mid\times x_i+q_j-mid\times p_j\ge 0\)

那么我们实际上就是要找一个\(i\)使得\(-x_i\times mid+y_i\)最大，同理找一个\(j\)使得\(-p_j\times mid+q_j\)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\(-a_ix+b_i\ge -a_jx+b_j, x\ge \frac{b_j-b_i}{a_j-a_i}\).

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\(x\ge \frac{b_j-b_i}{a_j-a_i}\), \(k\)不劣于\(j\)当且仅当\(x\le \frac{b_k-b_j}{a_k-a_j}\), 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\(O(n\log^4n)\)

(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\n",left);

	}

	return 0;

}

巴特西

BZOJ 2402 陶陶的难题II (树链剖分、线段树、凸包、分数规划)

代码

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