题目大意：一些成熟分布在第一象限中，现在要建造一个防线来保护他们，但是随着时间的推移，必须要舍弃一些城市，但是不会舍弃首都。问最短的防线需要多长。

思路：在每一个时刻求一个上凸包就是答案了。当然这样做时间复杂度就呵呵了。考虑一下动态维护凸包。因为只有上凸包，所以处理起来会相对方便。我们只需把在凸包中的点按照x坐标排序，然后二分一下把点插入凸包，然后左右用斜率维护一下，这样每次插点的时间复杂度大概是O(logn)。但是这样只能插点不能删点，所以离线处理一下，把删点转化为插点，最后倒着输出。

（我比较懒，然后为省去平衡树直接上set了，可恶的iterator不支持+1，写的这个DT。。。

CODE：

#include <set>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 200010
#define EPS 1e-10
using namespace std;

struct Operation{
	int flag,p;
	
	void Read() {
		scanf("%d",&flag);
		if(flag == 1)	scanf("%d",&p);
	}
}opt[MAX];
struct Point{
	int x,y;
	
	Point(int _ = 0,int __ = 0):x(_),y(__) {}
 	bool operator <(const Point &a)const {
		if(x == a.x)	return y < a.y;
		return x < a.x;
	}
	void Read() {
		scanf("%d%d",&x,&y);
	}
}point[MAX];
set<Point> convex;
set<Point>::iterator first,last;

int cnt,asks;
int x,y,n;
bool ban[MAX];
double ans[MAX];

double length;

inline double Calc(const Point &a,const Point &b)
{
	return sqrt((double)(a.x - b.x) * (a.x - b.x) + (double)(a.y - b.y) * (a.y - b.y));
}

inline double GetSlope(const Point &a,const Point &b)
{
	if(a.x == b.x)	return 1e15;
	return (double)(b.y - a.y) / (b.x - a.x);
}

inline void Insert(const Point &p)
{
	set<Point>::iterator it = convex.insert(p).first;
	set<Point>::iterator _it = --it; ++it;
	set<Point>::iterator __it = first;
	if(_it != first) {
		__it = --_it; ++_it;
	}
	set<Point>::iterator it_ = ++it; --it;
	set<Point>::iterator it__ = last;
	if(it_ != last) {
		it__ = ++it_; --it_;
	}
	if(GetSlope(*_it,*it) > GetSlope(*it,*it_))
		length -= Calc(*_it,*it_);
	else {
		convex.erase(it);
		return ;
	}
	while(_it != first && GetSlope(*__it,*_it) < GetSlope(*_it,*it)) {
		length -= Calc(*_it,*__it);
		convex.erase(_it);
		_it = __it--;
	}
	while(it_ != last && GetSlope(*it,*it_) < GetSlope(*it_,*it__)) {
		length -= Calc(*it_,*it__);
		convex.erase(it_);
		it_ = it__++;
	}
	length += Calc(*it,*_it) + Calc(*it,*it_);	
}

int main()
{
	cin >> n >> x >> y;
	first = convex.insert(Point(0,0)).first;
	convex.insert(Point(x,y));
	last = convex.insert(Point(n,0)).first;
	length = Calc(Point(x,y),Point(0,0)) + Calc(Point(x,y),Point(n,0));
	cin >> cnt;
	for(int i = 1; i <= cnt; ++i)
		point[i].Read();
	cin >> asks;
	for(int i = 1; i <= asks; ++i) {
		opt[i].Read();
		ban[opt[i].p] = true;
	}
	for(int i = 1; i <= cnt; ++i)
		if(!ban[i])
			Insert(point[i]);
	for(int i = asks; i; --i) {
		if(opt[i].flag == 1)
			Insert(point[opt[i].p]);
		else	ans[i] = length;
	}
	for(int i = 1; i <= asks; ++i) 
		if(fabs(ans[i]) > EPS)
			printf("%.2lf\n",ans[i]);
	return 0;
}

BZOJ 2300 HAOI 2011 防线修建 动态维护凸包