BZOJ3533 [Sdoi2014]向量集 【线段树 + 凸包 + 三分】

题目链接

BZOJ3533

题解

我们设询问的向量为\((x_0,y_0)\)，参与乘积的向量为\((x,y)\)

则有

\[\begin{aligned}
ans &= x_0x + y_0y \\
y &= -\frac{x_0}{y_0}x + \frac{ans}{y_0} \\
\end{aligned}
\]

所以向量集里的向量实际上可以对应到平面上一组点，我们用一个斜率固定的直线去经过这些点，使得斜率最大或最小

当\(y_0 > 0\)时，要求截距最大

当\(y_0 < 0\)时，要求截距最小

当\(y_0 = 0\)时，只用讨论\(x\)

这样看来，能产生贡献的点，一定是凸包上的点，前者是上凸包，后者是下凸包，\(y_0 = 0\)要求\(x\)的极值，上下凸包都可

所以我们的问题就转化为了如何快速求出区间的凸包

由于是求最优解，我们并不需要每次都对整个区间建一个凸包，分成若干个小凸包合并答案也是可以的

所以我们可以线段树维护区间凸包，当一个区间满的时候再建立凸包即可

由于每一层都是\(n\)个位置，所以总的复杂度是\(O(nlog^2n)\)的

询问的时候在每个凸包上三分，也是\(O(nlog^2n)\)的

有几点要注意的：

①INF要足够大

②整型三分的姿势

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define pb push_back
#define ls (u << 1)
#define rs (u << 1 | 1)
using namespace std;
const int maxn = 400005,maxm = 100005;
const LL INF = 9223372036854775807ll;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
struct point{
	LL x,y;
}p[maxn],t[maxn];
inline point operator -(const point& a,const point& b){
	return (point){a.x - b.x,a.y - b.y};
}
inline point operator +(const point& a,const point& b){
	return (point){a.x + b.x,a.y + b.y};
}
inline LL cross(const point& a,const point& b){
	return a.x * b.y - a.y * b.x;
}
inline LL operator *(const point& a,const point& b){
	return a.x * b.x + a.y * b.y;
}
inline double slope(const point& a,const point& b){
	if (a.x == b.x) return INF;
	return 1.0 * (a.y - b.y) / (a.x - b.x);
}
inline bool operator <(const point& a,const point& b){
	return a.x == b.x ? a.y < b.y : a.x < b.x;
}
inline bool operator ==(const point& a,const point& b){
	return a.x == b.x && a.y == b.y;
}
int n,now; char S;
LL lans;
inline int decode (int x ,LL lastans) {
	return S == 'E' ? x : (x ^ (lastans & 0x7fffffff));
}
vector<point> Tu[maxn << 2],Td[maxn << 2];
int top[maxn << 2],Top[maxn << 2];
point st[maxn];
void build(int u,int l,int r){
	int N = 0;
	for (int i = l; i <= r; i++) t[++N] = p[i];
	sort(t + 1,t + 1 + N);
	top[u] = -1;
	for (int i = 1; i <= N; i++){
		if (i > 1 && t[i] == t[i - 1]) continue;
		while (top[u] > 0 && cross(st[top[u]] - st[top[u] - 1],t[i] - st[top[u]]) >= 0)
			top[u]--;
		st[++top[u]] = t[i];
	}
	for (int i = 0; i <= top[u]; i++) Tu[u].pb(st[i]);
	Top[u] = -1;
	for (int i = 1; i <= N; i++){
		if (i > 1 && t[i] == t[i - 1]) continue;
		while (Top[u] > 0 && cross(st[Top[u]] - st[Top[u] - 1],t[i] - st[Top[u]]) <= 0)
			Top[u]--;
		st[++Top[u]] = t[i];
	}
	for (int i = 0; i <= Top[u]; i++) Td[u].pb(st[i]);
}
void insert(int u,int l,int r){
	if (l == r){Tu[u].pb(p[now]); Td[u].pb(p[now]); top[u] = Top[u] = 0; return;}
	if (now == r) build(u,l,r);
	int mid = l + r >> 1;
	if (mid >= now) insert(ls,l,mid);
	else insert(rs,mid + 1,r);
}
LL qmax(int u,point P){
	if (P.y >= 0){
		int l = 0,r = top[u],lmid,rmid;
		while (r - l >= 3){
			lmid = (l + l + r) / 3;
			rmid = (r + l + r) / 3;
			if (P * Tu[u][lmid] <= P * Tu[u][rmid]) l = lmid;
			else r = rmid;
		}
		LL ans = -INF;
		for (int i = l; i <= r; i++) ans = max(ans,P * Tu[u][i]);
		return ans;
	}
	else {
		int l = 0,r = Top[u],lmid,rmid;
		while (r - l >= 3){
			lmid = (l + l + r) / 3;
			rmid = (r + l + r) / 3;

			if (P * Td[u][lmid] <= P * Td[u][rmid]) l = lmid;
			else r = rmid;
		}
		LL ans = -INF;
		for (int i = l; i <= r; i++) ans = max(ans,P * Td[u][i]);
		return ans;
	}
}
LL query(int u,int l,int r,int L,int R,point a){
	if (l >= L && r <= R) return qmax(u,a);
	int mid = l + r >> 1;
	if (mid >= R) return query(ls,l,mid,L,R,a);
	if (mid < L) return query(rs,mid + 1,r,L,R,a);
	return max(query(ls,l,mid,L,R,a),query(rs,mid + 1,r,L,R,a));
}
int main(){
	n = read(); scanf("%c",&S);
	char opt; int x,y,l,r;
	for (int i = 1; i <= n; i++){
		opt = getchar(); while (opt != 'A' && opt != 'Q') opt = getchar();
		x = decode(read(),lans);
		y = decode(read(),lans);
		if (opt == 'A'){
			p[++now] = (point){x,y};
			insert(1,1,n);
		}
		else {
			l = decode(read(),lans);
			r = decode(read(),lans);
			lans = query(1,1,n,l,r,(point){x,y});
			printf("%lld\n",lans);
		}
	}
	return 0;
}