https://www.luogu.org/problemnew/show/2521
题意:维护一个上凸包:删点,查询周长
很容易想到把问题转换为离线:先读入全部操作,记录下最后剩下的点,倒着加点来维护凸包,同时也倒着做询问。
然后问题就变成了怎么维护加点的操作,这题其实只要维护上半个凸包(其实也有一点启发性了吧),用set存凸包的点集,对于要加的点往左右两边一直把不行的点删掉就好了,因为一个点最多被删一次所以加上set的$log$实际复杂度是$O(nlog n)$的而不是$O(n^2)$。
如果要维护整个凸包就根据y来记录一下两边的位置根据上下凸包分类讨论一下来维护。
#include<cstdio> #include<cmath> #include<set> #include<algorithm> using namespace std; const int N=100005; struct q { double ans; int op,idx; }ask[N<<1]; struct Point { double x,y; Point(double x=0,double y=0):x(x),y(y){} }p[N]; inline bool operator <(Point a,Point b) { if(a.x==b.x)return a.y<b.y; return a.x<b.x; } inline Point operator - (Point a,Point b) { return Point(a.x-b.x,a.y-b.y); } inline double cross(Point a,Point b) { return a.x*b.y-a.y*b.x; } inline double dot(Point a,Point b) { return a.x*b.x+a.y*b.y; } inline double lenght(Point a) { return sqrt(dot(a,a)); } double ans; set<Point>s; bool mark[N]; inline void ins(Point x) { set<Point>::iterator r=s.lower_bound(x); set<Point>::iterator l=r;l--; if(cross(*r-*l,x-*l)<0)return; ans-=lenght(*r-*l); s.insert(x); while(1) { set<Point>::iterator it=r++; if(r==s.end())break; if(cross(*r-x,*it-x)>0)break; ans-=lenght(*it-*r); s.erase(*it); } while(1) { if(l==s.begin())break; set<Point>::iterator it=l--; if(cross(*l-x,*it-x)<0)break; ans-=lenght(*it-*l); s.erase(*it); } l=r=s.find(x);l--;r++; ans+=lenght(x-*l)+lenght(x-*r); } int main() { int m,q;double x,y,n; scanf("%lf%lf%lf",&n,&x,&y); s.insert(Point(0,0));s.insert(Point(n,0));s.insert(Point(x,y)); ans+=lenght(Point(x,y))+lenght(Point(x-n,y)); scanf("%d",&m); for(register int i=1;i<=m;i++)scanf("%lf%lf",&p[i].x,&p[i].y); scanf("%d",&q); for(register int i=1;i<=q;i++) { scanf("%d",&ask[i].op); if(ask[i].op==1) scanf("%d",&ask[i].idx),mark[ask[i].idx]=1; } for(register int i=1;i<=m;i++)if(!mark[i])ins(p[i]); for(register int i=q;i>=1;i--) { if(ask[i].op==2)ask[i].ans=ans; else ins(p[ask[i].idx]); } for(register int i=1;i<=q;i++)if(ask[i].op==2)printf("%.2lf\n",ask[i].ans); return 0; }