【bzoj 2300】: [HAOI2011]防线修建

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http://www.lydsy.com/JudgeOnline/problem.php?id=2300

离线+set维护凸包

第一次码这个，还跑得飞快。。。

#include <cstdio>#include <cstring>#include <cstdlib>#include <iostream>#include <cmath>#include <algorithm>#include <set>using namespace std;#define rep(i,l,r) for(int i=(l),_=(r);i<=_;i++)#define per(i,r,l) for(int i=(r),_=(l);i>=_;i--)#define MS(arr,x) memset(arr,x,sizeof(arr))#define INE(i,u) for(int i=head[u];~i;i=e[i].next)#define LL long longinline const int read(){int r=0,k=1;char c=getchar();for(;c<'0'||c>'9';c=getchar())if(c=='-')k=-1;for(;c>='0'&&c<='9';c=getchar())r=r*10+c-'0';return k*r;}////////////////////////////////////////////////int n,m,q;struct Point{int x,y;Point(){}Point(int x,int y):x(x),y(y){}Point operator-(const Point &b)const{return Point(x-b.x,y-b.y);}int operator^(const Point &b)const{return x*b.y-y*b.x;}bool operator<(const Point &b)const{return x!=b.x?x<b.x:y<b.y;}}M,a[100010];struct ques{int type,x;}b[200010];bool del[100010];set<Point>S;double ans[200010],nowans;int tot;////////////////////////////////////////////////double sqr(double x){return x*x;}double dis(Point A,Point B){return sqrt(sqr(A.x-B.x)+sqr(A.y-B.y));}void insert(Point P){set<Point>::iterator L,R=S.lower_bound(P),tmp;L=R; L--;if((*R-P^*L-P)>=0) return;nowans-=dis(*L,*R);while(L->x!=0){tmp=L; tmp--;if((P-*L^*tmp-*L)<0) break;nowans-=dis(*L,*tmp);S.erase(*L);L=tmp;}while(R->x!=n){tmp=R; tmp++;if((*tmp-*R^P-*R)<0) break;nowans-=dis(*R,*tmp);S.erase(*R);R=tmp;}S.insert(P);nowans+=dis(*L,P);nowans+=dis(P,*R);}////////////////////////////////////////////////void input(){    n=read();    M.x=read(); M.y=read();    rep(i,1,m=read()) a[i].x=read(),a[i].y=read();    rep(i,1,q=read())    {    b[i].type=read();    if(b[i].type==1)    {    b[i].x=read();    del[b[i].x]=1;    }    }    S.insert(Point(0,0));    S.insert(Point(n,0));    S.insert(M);    nowans=dis(Point(0,0),M)+dis(M,Point(n,0));    rep(i,1,m) if(!del[i]) insert(a[i]);}void solve(){per(i,q,1){if(b[i].type==1){insert(a[b[i].x]);}else ans[++tot]=nowans;}per(i,tot,1) printf("%.2f\n",ans[i]);}////////////////////////////////////////////////int main(){    freopen("_.in","r",stdin); freopen("_.out","w",stdout);    input(),solve();    return 0;}

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