【bzoj2300】[HAOI2011]防线修建 凸包+set

离线，将删除操作变成插入操作
每次插入一个点，要求动态维护凸包
set维护，学着用指针迭代器吧

插入一个点，依次把两边不符合要求的点删除，用迭代器直接++和--就可以了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<set>
#define maxn 100010

using namespace std;

struct yts
{
	int x,y;
}a[maxn];

struct yts1
{
	int op,x;
	double ans;
}p[200010];

int n,m,T;
double ans;
set<yts> q;
bool vis[maxn];

yts operator-(yts x,yts y)
{
	yts ans;
	ans.x=x.x-y.x;ans.y=x.y-y.y;
	return ans;
}

double operator*(yts x,yts y)
{
	return x.x*y.y-x.y*y.x;
}

bool operator<(yts x,yts y)
{
	return x.x<y.x || (x.x==y.x && x.y<y.y);
}

double dis(yts x,yts y)
{
	return sqrt((double)(y.x-x.x)*(y.x-x.x)+(y.y-x.y)*(y.y-x.y));
}

void insert(int x,int y)
{
	yts p;
	p.x=x;p.y=y;
	set<yts>::iterator r=q.lower_bound(p),l=r,t;
	l--;
	if ((*r-*l)*(p-*l)<0) return;
	ans-=dis(*l,*r);
	while (1)
	{
		t=r;r++;
		if (r==q.end()) break;
		if ((*r-p)*(*t-p)>0) break;
		ans-=dis(*t,*r);
		q.erase(t);
	}
	while (l!=q.begin())
	{
		t=l;l--;
		if ((*t-p)*(*l-p)>0) break;
		ans-=dis(*l,*t);
		q.erase(t);
	}
	q.insert(p);
	l=r=t=q.find(p);
	l--;r++;
	ans+=dis(*l,*t)+dis(*t,*r);
}

int main()
{
	scanf("%d",&n);
	yts p1,p2,p3;int x,y;
	p1.x=0;p1.y=0;q.insert(p1);
	p2.x=n;p2.y=0;q.insert(p2);
	scanf("%d%d",&x,&y);
	p3.x=x;p3.y=y;q.insert(p3);
	ans=dis(p1,p3)+dis(p2,p3);
	scanf("%d",&m);
	for (int i=1;i<=m;i++) scanf("%d%d",&a[i].x,&a[i].y);
	memset(vis,0,sizeof(vis));
	scanf("%d",&T);
	for (int i=1;i<=T;i++)
	{
		scanf("%d",&p[i].op);
		if (p[i].op==1) {scanf("%d",&p[i].x);vis[p[i].x]=1;}
	}
	for (int i=1;i<=m;i++) if (!vis[i]) insert(a[i].x,a[i].y);
	for (int i=T;i>=1;i--)
	{
		if (p[i].op==1) insert(a[p[i].x].x,a[p[i].x].y);
		else p[i].ans=ans;
	}
	for (int i=1;i<=T;i++) if (p[i].op==2) printf("%.2lf\n",p[i].ans);
	return 0;
}