【BZOJ】4311: 向量（线段树分治板子题）

题解

我们可以根据点积的定义，垂直于原点到给定点构成的直线作一条直线，从正无穷往下平移，第一个碰到的点就是答案

像什么，上凸壳哇

可是……动态维护上凸壳？

我们可以离线，计算每个点能造成贡献的一个询问区间[l,r]表示这个点在第l个询问和第r个询问之间存在，按照每个点的横坐标大小顺序插入线段树，我们就可以类似斜率优化构造出凸包

对于所有询问，我们可以给它们按极角排序，然后遍历线段树，如果按照极角排序，那么垂直于他们的直线斜率递减，最优点也右移

实现的方法就是一边遍历线段树，一边归并排序，每一层按照极角序遍历这个凸壳就好

代码

#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define MAXN 200005
#define RG register
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;char c = getchar();T f = 1;
	    while(c < '0' || c > '9') {
			if(c == '-') f = -1;
			c = getchar();
	    }
	    while(c >= '0' && c <= '9') {
		res = res * 10 + c - '0';
		c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {putchar('-');x = -x;}
    if(x >= 10) {
		out(x / 10);
    }
    putchar('0' + x % 10);
}
struct Point {
	int64 x,y;
	Point(){}
	Point(int64 _x,int64 _y) {
		x = _x;y = _y;
	}
	friend Point operator + (const Point &a,const Point &b) {
		return Point(a.x + b.x,a.y + b.y);
	}
	friend Point operator - (const Point &a,const Point &b) {
		return Point(a.x - b.x,a.y - b.y);
	}
	friend int64 operator * (const Point &a,const Point &b) {
		return a.x * b.y - a.y * b.x;
	}
	friend int64 dot(const Point &a,const Point &b) {
		return a.x * b.x + a.y * b.y;
	}
	friend bool operator < (const Point &a,const Point &b) {
		return a.x < b.x;
	}
};
struct Inode {
	Point a;int l,r;
	friend bool operator < (const Inode &s,const Inode &t) {
		return s.a < t.a;
	}
}Ins[MAXN];
struct Qnode {
	Point a;int id;
}Qry[MAXN],tmp[MAXN];
vector<Point> tr[MAXN * 4];
int N,cntI,cntQ,st[MAXN * 4];
int64 ans[MAXN];
void Insert(int u,int L,int R,int l,int r,Point a) {
	if(L == l && R == r) {
		int s = tr[u].size() - 1;
		while(s > 0) {
			if((tr[u][s] - tr[u][s - 1]) * (a - tr[u][s - 1]) >= 0) {
				tr[u].pop_back();
			}
			else break;
			--s;
		}
		tr[u].pb(a);
		return ;
	}
	int mid = (L + R) >> 1;
	if(r <= mid) Insert(u << 1,L,mid,l,r,a);
	else if(l > mid) Insert(u << 1 | 1,mid + 1,R,l,r,a);
	else {
		Insert(u << 1,L,mid,l,mid,a);
		Insert(u << 1 | 1,mid + 1,R,mid + 1,r,a);
	}
}
void Init() {
	read(N);
	int t,id;int64 x,y;
	for(int i = 1 ; i <= N ; ++i) {
		read(t);
		if(t == 1) {
			read(x);read(y);
			Ins[++cntI] = (Inode){Point(x,y),cntQ + 1,-1};
		}
		else if(t == 3) {
			read(x);read(y);++cntQ;
			Qry[cntQ] = (Qnode){Point(x,y),cntQ};
		}
		else {
			read(id);
			Ins[id].r = cntQ;
		}
	}
	for(int i = 1 ; i <= cntI ; ++i) {
		if(Ins[i].r == -1) Ins[i].r = cntQ;
	}
	sort(Ins + 1,Ins + cntI + 1);
	for(int i = 1 ; i <= cntI ; ++i) {
		if(Ins[i].l <= Ins[i].r) {
			Insert(1,1,cntQ,Ins[i].l,Ins[i].r,Ins[i].a);
		}
	}
}
void Solve(int u,int L,int R) {
	int mid = (L + R) >> 1;
	if(L != R) {
		Solve(u << 1,L,mid);Solve(u << 1 | 1,mid + 1,R);
		int tl = L,tr = mid + 1,p = L;
		while(tl <= mid && tr <= R) {
			if(Qry[tr].a * Qry[tl].a >= 0) tmp[p++] = Qry[tl++];
			else tmp[p++] = Qry[tr++];
		}
		while(tl <= mid) tmp[p++] = Qry[tl++];
		while(tr <= R) tmp[p++] = Qry[tr++];
		for(int i = L ; i <= R ; ++i) Qry[i] = tmp[i];
	}
	int s = tr[u].size() - 1;
	if(s != -1) {
		for(int i = L ; i <= R ; ++i) {
			while(st[u] < s) {
				if(dot(tr[u][st[u]],Qry[i].a) <= dot(tr[u][st[u] + 1],Qry[i].a)) {
					++st[u];
				}
				else break;
			}
			ans[Qry[i].id] = max(ans[Qry[i].id],dot(Qry[i].a,tr[u][st[u]])); 
		}
	}
	
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve(1,1,cntQ);
    for(int i = 1 ; i <= cntQ ; ++i) out(ans[i]),enter;
    return 0;
}