latex預處理

\( \newcommand{\ord}[1]{\mathcal{O}\left(#1\right)} \newcommand{\abs}[1]{\lvert #1 \rvert} \newcommand{\floor}[1]{\lfloor #1 \rfloor} \newcommand{\ceil}[1]{\lceil #1 \rceil} \newcommand{\opord}{\operatorname{\mathcal{O}}} \newcommand{\argmax}{\operatorname{arg\,max}} \newcommand{\str}[1]{\texttt{"#1"}} \)

2016年2月20日 星期六

[ IOICAMP2016 ] 動態曼哈頓最短距離

題目:
因為IOICAMP的judge是暫時性的，所以有人備份了題目
http://codingsimplifylife.blogspot.tw/2016/02/ioi-camp-judge-37_4.html

解法:
CDQ分治不會寫，所以就用老派的kd tree去解吧，這裡提供兩種寫法:
  • 第一種寫法是動態的將點進行插入，因為必須要做到動態的插入操作，而一般kd tree不能用旋轉的方式來平衡，所以利用替罪羊樹的概念來平衡。(3.10s)
  •  第二種寫法是把所有操作讀入，把所有要插入的點先建成一顆kd tree，接著倒著作回來，如果遇到插入操作就把點從kd tree裡刪除，但是刪除的速度很慢，所以是壓線過的。(7.16)
兩種寫法如果再kdt.clear()時不做delete操作只把root=0的話會加快約2秒的時間(3.02s跟3.73s)

第一種作法code:
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#include<stdio.h>
#include<limits.h>
//using namespace std;
#ifndef SUNMOON_DYNEMIC_KD_TREE
#define SUNMOON_DYNEMIC_KD_TREE
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
template<typename T,size_t kd>//kd表示有幾個維度
class kd_tree{
    public:
        struct point{
            T d[kd];
            inline T dist(const point &x)const{
                T ret=0;
                for(size_t i=0;i<kd;++i)ret+=std::abs(d[i]-x.d[i]);
                return ret;
            }
            inline bool operator<(const point &b)const{
                return d[0]<b.d[0];
            }
        };
    private:
        struct node{
            node *l,*r;
            point pid;
            int s;
            node(const point &p):l(0),r(0),pid(p),s(1){}
            inline void up(){
                s=(l?l->s:0)+1+(r?r->s:0);
            }
        }*root;
        const double alpha,loga;
        const T INF;//記得要給INF，表示極大值
        std::vector<node*> A;
        int qM;
        std::priority_queue<std::pair<T,point > >pQ;
        struct __cmp{
            int sort_id;
            inline bool operator()(const node*x,const node*y)const{
                return x->pid.d[sort_id]<y->pid.d[sort_id];
            }
        }cmp;
        void clear(node *o){
            if(!o)return;
            clear(o->l);
            clear(o->r);
            delete o;
        }
        inline int size(node *o){
            return o?o->s:0;
        }
        node* build(int k,int l,int r){
            if(l>r)return 0;
            if(k==kd)k=0;
            int mid=(l+r)/2;
            cmp.sort_id=k;
            std::nth_element(A.begin()+l,A.begin()+mid,A.begin()+r+1,cmp);
            node *ret=A[mid];
            ret->l=build(k+1,l,mid-1);
            ret->r=build(k+1,mid+1,r);
            ret->up();
            return ret;
        }
        inline bool isbad(node*o){
            return size(o->l)>alpha*o->s||size(o->r)>alpha*o->s;
        }
        void flatten(node *u,typename std::vector<node*>::iterator &it){
            if(!u)return;
            flatten(u->l,it);
            *it=u;
            flatten(u->r,++it);
        }
        bool insert(node*&u,int k,const point &x,int dep){
            if(!u){
                u=new node(x);
                return dep<=0;
            }
            ++u->s;
            if(insert(x.d[k]<u->pid.d[k]?u->l:u->r,(k+1)%kd,x,dep-1)){
                if(!isbad(u))return 1;
                if((int)A.size()<u->s)A.resize(u->s);
                typename std::vector<node*>::iterator it=A.begin();
                flatten(u,it);
                u=build(k,0,u->s-1);
            }
            return 0;
        }
        inline int heuristic(const int h[])const{
            int ret=0;
            for(size_t i=0;i<kd;++i)ret+=h[i];
            return ret;
        }
        void nearest(node *u,int k,const point &x,T *h,T &mndist){
            if(u==0||heuristic(h)>=mndist)return;
            point now=u->pid;
            int dist=u->pid.dist(x),old=h[k];
            /*mndist=std::min(mndist,dist);*/
            if(dist<mndist){
                pQ.push(std::make_pair(dist,u->pid));
                if((int)pQ.size()==qM+1){
                    mndist=pQ.top().first,pQ.pop();
                }
            }
            if(x.d[k]<u->pid.d[k]){
                nearest(u->l,(k+1)%kd,x,h,mndist);
                h[k]=abs(x.d[k]-u->pid.d[k]);
                nearest(u->r,(k+1)%kd,x,h,mndist);
            }else{
                nearest(u->r,(k+1)%kd,x,h,mndist);
                h[k]=abs(x.d[k]-u->pid.d[k]);
                nearest(u->l,(k+1)%kd,x,h,mndist);
            }
            h[k]=old;
        }
        std::vector<point>in_range;
        void range(node *u,int k,const point&mi,const point&ma){
            if(!u)return;
            bool is=1;
            for(int i=0;i<kd;++i)
                if(u->pid.d[i]<mi.d[i]||ma.d[i]<u->pid.d[i]){
                    is=0;break;
                }
            if(is)in_range.push_back(u->pid);
            if(mi.d[k]<=u->pid.d[k])range(u->l,(k+1)%kd,mi,ma);
            if(mi.d[k]>=u->pid.d[k])range(u->r,(k+1)%kd,mi,ma);
        }
    public:
        kd_tree(const T &INF,double a=0.75):alpha(a),loga(log2(1.0/a)),INF(INF){}
        inline void clear(){
            clear(root),root=0;
        }
        inline void build(int n,const point *p){
            clear(root),A.resize(n);
            for(int i=0;i<n;++i)A[i]=new node(p[i]);
            root=build(0,0,n-1);
        }
        inline void insert(const point &x){
            insert(root,0,x,std::__lg(size(root))/loga);
        }
        inline T nearest(const point &x,int k){
            qM=k;
            T mndist=INF,h[kd]={};
            nearest(root,0,x,h,mndist);
            mndist=pQ.top().first;
            pQ=std::priority_queue<std::pair<T,point > >();
            return mndist;/*回傳離x第k近的點的距離*/
        }
        inline const std::vector<point> &range(const point&mi,const point&ma){
            in_range.clear();
            range(root,0,mi,ma);
            return in_range;/*回傳介於mi到ma之間的點vector*/
        }
        inline int size(){return root?root->s:0;}
};
#endif
kd_tree<int,2> kdt(INT_MAX);
int t,n,a;
kd_tree<int,2>::point x;
int main(){
    scanf("%d",&t);
    while(t--){
        kdt.clear();
        scanf("%d",&n);
        while(n--){
            scanf("%d%d%d",&a,&x.d[0],&x.d[1]);
            if(a)printf("%d\n",kdt.nearest(x,1));
            else kdt.insert(x);
        }
    }
    return 0;
}

第二種做法code:
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#include<stdio.h>
#include<limits.h>
#include<assert.h>
//using namespace std; 
#ifndef SUNMOON_DYNEMIC_KD_TREE
#define SUNMOON_DYNEMIC_KD_TREE
#include<algorithm>
#include<vector>
template<typename T,size_t kd>
class kd_tree{
    public:
        struct point{
            T d[kd];
            inline T dist(const point &x)const{
                T ret=0;
                for(size_t i=0;i<kd;++i)ret+=std::abs(d[i]-x.d[i]);
                return ret;
            }
            inline bool operator<(const point &b)const{
                return d[0]<b.d[0];
            }
        };
    private:
        struct node{
            node *l,*r;
            point pid;
            node(const point &p):l(0),r(0),pid(p){}
        }*root;
        const T INF;
        std::vector<node*> A;
        int s;
        struct __cmp{
            int sort_id;
            inline bool operator()(const node*x,const node*y)const{
                return x->pid.d[sort_id]<y->pid.d[sort_id];
            }
        }cmp;
        void clear(node *o){
            if(!o)return;
            clear(o->l);
            clear(o->r);
            delete o;
        }
        node* build(int k,int l,int r){
            if(l>r)return 0;
            if(k==kd)k=0;
            int mid=(l+r)/2;
            cmp.sort_id=k;
            std::nth_element(A.begin()+l,A.begin()+mid,A.begin()+r+1,cmp);
            node *ret=A[mid];
            ret->l=build(k+1,l,mid-1);
            ret->r=build(k+1,mid+1,r);
            return ret;
        }
        inline int heuristic(const int h[])const{
            int ret=0;
            for(size_t i=0;i<kd;++i)ret+=h[i];
            return ret;
        }
        node **mnp;
        int mnk;
        void findmin(node*&o,int d,int k){
            if(!o)return;
            if(!mnp||o->pid.d[d]<(*mnp)->pid.d[d]){
                mnp=&o;
                mnk=k;
            }
            findmin(o->l,d,(k+1)%kd);
            if(d==k)return;
            findmin(o->r,d,(k+1)%kd);
        }
        void nearest_for_erase(node *&u,int k,const point &x,T *h,T &mndist){
            if(u==0||heuristic(h)>=mndist)return;
            point now=u->pid;
            int dist=u->pid.dist(x),old=h[k];
            if(dist<mndist){
                mnp=&u;
                mnk=k;
                if(!(mndist=dist))return;
            }
            if(x.d[k]<u->pid.d[k]){
                nearest_for_erase(u->l,(k+1)%kd,x,h,mndist);
                h[k]=abs(x.d[k]-u->pid.d[k]);
                nearest_for_erase(u->r,(k+1)%kd,x,h,mndist);
            }else{
                nearest_for_erase(u->r,(k+1)%kd,x,h,mndist);
                h[k]=abs(x.d[k]-u->pid.d[k]);
                nearest_for_erase(u->l,(k+1)%kd,x,h,mndist);
            }
            h[k]=old;
        }
    public:
        kd_tree(const T &INF):INF(INF),s(0){}
        inline void clear(){
            clear(root),root=0;
        }
        inline void build(int n,const point *p){
            clear(root),A.resize(s=n);
            for(int i=0;i<n;++i)A[i]=new node(p[i]);
            root=build(0,0,n-1);
        }
        inline bool erase(point p){
            T mndist=1,h[kd]={};
            nearest_for_erase(root,0,p,h,mndist);
            if(mndist)return 0;
            for(node **o=mnp;;){
                if((*o)->r);
                else if((*o)->l){
                    (*o)->r=(*o)->l;
                    (*o)->l=0;
                }else{
                    delete *o;
                    (*o)=0;
                    --s;
                    return 1;
                }
                mnp=0;
                findmin((*o)->r,mnk,(mnk+1)%kd);
                (*o)->pid=(*mnp)->pid;
                o=mnp;
            }
        }
        inline T nearest(const point &x){
            T mndist=INF,h[kd]={};
            nearest_for_erase(root,0,x,h,mndist);
            return mndist;/*回傳離x最近的點的距離*/
        }
        inline int size(){return s;}
};
#endif
kd_tree<int,2> kdt(INT_MAX);
int t,n;
kd_tree<int,2>::point x[200005],in[200005];
int ans[200005],a[200005],top;
int main(){
    scanf("%d",&t);
    while(t--){
        kdt.clear();
        scanf("%d",&n);
        top=0;
        for(int i=0;i<n;++i){
            scanf("%d%d%d",&a[i],&x[i].d[0],&x[i].d[1]);
            if(!a[i])in[top++]=x[i];
        }
        kdt.build(top,in);
        top=0;
        while(n--){
            if(a[n])ans[top++]=kdt.nearest(x[n]);
            else assert(kdt.erase(x[n]));
        }
        while(top--)printf("%d\n",ans[top]);
    }
    return 0;
}

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