BZOJ 2300 HAOI2011 防线修建 平衡树维护凸包

题目大意：给定初始三个点(0,0)(n,0)和(x,y)，以及若干其它点，q次询问，每次删除一个点或求一次上凸包长度

平衡树维护凸包……和cash同样的思路为何我时隔2个月后的代码整整短了150行……

平衡树最好用set 链表的每个点存set的迭代器 直接erase很方便

这个题x值的下限和上限都是给定的 所以没有边界讨论 真赞~

一气呵成写出来 调了一小时 尼玛加点我居然加反了0.0

#include<set>#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#define M 100100#define INF 1e10using namespace std;typedef pair<double,double> point;struct list;typedef set< pair<double,list*> >::iterator sit;struct list{list *next,*last;point p;sit it;}head,tail;set< pair<double,list*> >s;int n,m,q;int queries[M<<1];double stack[M];int top;point a[M];bool v[M];double len;double Distance(const point &p1,const point &p2){return sqrt( (p1.first-p2.first)*(p1.first-p2.first) + (p1.second-p2.second)*(p1.second-p2.second) );}double Get_Slope(const point &p1,const point &p2){if(p1.first==p2.first)return p1.second<p2.second?INF:-INF;return (p1.second-p2.second)/(p1.first-p2.first);}void Insert(const point &p){list *secc=s.lower_bound( make_pair(p.first,(list*)0x0) )->second;list *pred=secc->last;if( Get_Slope(pred->p,p)<Get_Slope(p,secc->p) )return ;len-=Distance(pred->p,secc->p);while( pred!=&head && Get_Slope(pred->last->p,pred->p)<Get_Slope(pred->p,p) )len-=Distance(pred->last->p,pred->p),s.erase(pred->it),pred=pred->last;while( secc!=&tail && Get_Slope(p,secc->p)<Get_Slope(secc->p,secc->next->p) )len-=Distance(secc->p,secc->next->p),s.erase(secc->it),secc=secc->next;len+=Distance(pred->p,p);len+=Distance(p,secc->p);list *temp=pred->next=secc->last=new list;temp->last=pred;temp->next=secc;temp->p=p;temp->it=s.insert( make_pair(p.first,temp) ).first;}int main(){int i,x,y,p;cin>>m>>x>>y>>n;for(i=1;i<=n;i++)scanf("%lf%lf",&a[i].first,&a[i].second);cin>>q;for(i=1;i<=q;i++){scanf("%d",&p);if(p==2)queries[i]=-1;elsescanf("%d",&queries[i]),v[queries[i]]=1;}head.next=&tail;tail.last=&head;head.p=point(0,0);head.it=s.insert( make_pair(0,&head) ).first;tail.p=point(m,0);tail.it=s.insert( make_pair(m,&tail) ).first;len=m;Insert( point(x,y) );for(i=1;i<=n;i++)if(!v[i])Insert(a[i]);for(i=q;i;i--){if(queries[i]==-1)stack[++top]=len;elseInsert(a[queries[i]]);}while(top)printf("%.2lf\n",stack[top--]);}