BZOJ 2300 动态凸包

最新推荐文章于 2021-01-22 16:24:13 发布

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本文介绍了一个简单的算法练习，旨在通过维护凸包的上凸性来提升编程技巧，适合准备参加NOI2007比赛的选手。

很简单的东西，维护凸包的上凸性而已，无聊来练练手，准备些NOI 2007 cash

#include <cstdio>
#include <cmath>
#include <set>

using namespace std;

const int MAXN = 300001;
const int MAXQ = 300001; 

struct Point
{
	bool flag;
	double x,y;
	bool operator < (const Point &rhs) const
	{
		return x < rhs.x || (x == rhs.x && y < rhs.y);
	}
	Point (){}
	Point (double _x,double _y):x(_x),y(_y){flag = false;}
};

Point operator - (Point A,Point B)
{
	return (Point){A.x-B.x,A.y-B.y};
}

double Cross(Point A,Point B)
{
	return A.x*B.y-A.y*B.x;
}

Point P[MAXN];
set <Point> S;
set <Point> :: iterator it1,it2;

struct Q
{
	int oper,id;
};

Q Ques[MAXQ];

int n,q;
double m,x,y;
double ans;
double Ans[MAXQ];

inline double dist(double x1,double y1,double x2,double y2)
{
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

bool judge(Point p)
{
	it1 = it2 = S.lower_bound(p);
	it1--;
	if (Cross(*it2-*it1,p-*it1) < 0) return true;
	return false;
}

void insert(Point p)
{
	if (judge(p)) return;
	it1 = it2 = S.lower_bound(p);
	it1--;
	ans -= dist(it1->x,it1->y,it2->x,it2->y);
	it1 = it2 = --S.lower_bound(p);
	it2--;
	while (true)//it1 != S.begin())
	{
		if (Cross(p-*it2,*it1-*it2) < 0) 
		{
			ans -= dist(it1->x,it1->y,it2->x,it2->y);
			S.erase(it1);
		}
		else break;
		it1 = it2;
		if (it2 != S.begin()) it2--;
	}
	it1 = it2 = S.lower_bound(p);
	it2++;
	while (true)//it2 != S.end())
	{
		if (Cross(p-*it2,*it1-*it2) > 0)
		{
			ans -= dist(it1->x,it1->y,it2->x,it2->y);
			S.erase(it1);
		}
		else break;
		it1 = it2;
		it2++;
	}
	S.insert(p);
	it1 = it2 = S.find(p);
	it1++, it2--;
	ans += dist(p.x,p.y,it1->x,it1->y)+dist(p.x,p.y,it2->x,it2->y);
}

int main()
{
	scanf("%lf%lf%lf",&m,&x,&y);
	S.insert((Point){0,0});
	S.insert((Point){m,0});
	S.insert((Point){x,y});
	ans = dist(0,0,x,y)+dist(x,y,m,0);
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%lf%lf",&P[i].x,&P[i].y);
	scanf("%d",&q);
	for (int i=1;i<=q;i++) 
	{
		scanf("%d",&Ques[i].oper);
		if (Ques[i].oper == 1) scanf("%d",&Ques[i].id), P[Ques[i].id].flag = true;
	}
	for (int i=1;i<=n;i++) if (!P[i].flag) insert(P[i]);
	for (int i=q;i>=1;i--)
	{
		if (Ques[i].oper == 1) insert(P[Ques[i].id]);
		else Ans[i] = ans;
	}
	for (int i=1;i<=q;i++) if (Ques[i].oper == 2) printf("%.2lf\n",Ans[i]);
	return 0;
}