"Kd-tree" bzoj3290 Theresa and data structure

Offline all operations, the establishment of all possible points kd-tree,add equivalent to the weight of +1,cancel equivalent weight-1.

The modification operation is to record the FA for each point on the kd-tree and modify it from the bottom up.

Optimization: If the sumv==0 of a rectangle box is not entered. The area of the rectangle is recorded with only "meaningful" points (with a weight of 0 regardless of the value).

#include <cstdio> #include <algorithm> #include <stack> #include <cstring>using namespace std;    int f,c;inline void R (int &x) {c=0;f=1; for (; c< ' 0 ' | | C> ' 9 ';    C=getchar ()) if (c== '-') f=-1; for (x=0; c>= ' 0 ' &&c<= ' 9 ';    C=getchar ()) (x*=10) + = (c ' 0 '); X*=f;}    inline void P (int x) {if (x<10) putchar (x+ ' 0 '); Else{p (X/10);p Utchar (x%10+ ' 0 ');}} stack<int>zhan; #define N 100001#define KD 3int dn,n,root,m,qp[2][kd],idn;struct node{int ch[2],w,minn[kd],maxx[    Kd],p[kd],sumv,id;        void Init () {sumv=w;      for (int i=0;i<kd;++i) minn[i]=maxx[i]=p[i]; }}t[n];bool operator < (const node &a,const node &b) {return A.P[DN] < B.P[DN];}    inline void pushup (const int &rt) {T[RT].SUMV=T[RT].W; for (int i=0;i<2;++i) if (t[rt].ch[i]/* && t[t[rt].ch[i]].sumv*/) {t[rt].sumv+=t[t[rt].ch[i          ]].SUMV; for (int j=0;j<kd;++j) {t[rt].minn[j]=min (T[RT].MINN[J],T[T[RT].CH[I]].MINN[J]);            T[rt].maxx[j]=max (T[rt].maxx[j],t[t[rt].ch[i]].maxx[j]);    }}}int buildtree (int l=1,int r=n,int d=0) {dn=d;    int m= (l+r>>1);    Nth_element (t+l,t+m,t+r+1); T[M].    Init ();    if (l!=m) T[m].ch[0]=buildtree (L,m-1, (d+1)%KD);    if (m!=r) T[m].ch[1]=buildtree (M+1,r, (d+1)%KD);    Pushup (m); return m;} inline bool Inside (const int &o) {for (int i=0;i<kd;++i) if (Qp[0][i] > T[o].p[i] | |    T[o].p[i] > Qp[1][i]) return 0; return 1;} inline bool Allinside (const int &o) {for (int i=0;i<kd;++i) if (Qp[0][i] > T[o].minn[i] | |    T[o].maxx[i] > Qp[1][i]) return 0; return 1;} inline bool Cross (const int &o) {for (int i=0;i<kd;++i) if (Qp[0][i] > T[o].maxx[i] | |    T[o].minn[i] > Qp[1][i]) return 0; return 1;}    int ans;inline void Query (int rt=root) {if (Inside (RT)) ANS+=T[RT].W; for (int i=0;i<2;++i) if (T[rt].ch[i] && cross (T[rt].ch[i]) {if (Allinside (T[rt].ch[i])) ANS+=T[T[RT].CH[I]].SUMV;        else if (T[T[RT].CH[I]].SUMV) Query (T[rt].ch[i]);    }}int Val;char op[n][7];int dian[n][kd],rs[n],ids[n],ma[n],fa[n];void Update () {int u=ma[idn];    T[u].w+=val;    T[u].sumv+=val;    U=fa[u];        while (u) {t[u].sumv=t[u].w+t[t[u].ch[0]].sumv+t[t[u].ch[1]].sumv;//pushup (U);      U=fa[u];    }}int Main () {//Freopen ("Theresa9.in", "R", stdin);//Freopen ("Bzoj3290.out", "w", stdout);    R (n); for (int i=1;i<=n;++i) {R (t[i].p[0]); R (t[i].p[1]);        R (t[i].p[2]);        T[i].id=i;      T[i].w=1;    } R (M);        for (int i=1;i<=m;++i) {scanf ("%s", Op[i]);            if (op[i][0]== ' A ') {++n; R (T[n].p[0]); R (t[n].p[1]);            R (t[n].p[2]);            T[n].id=n;            Ids[i]=n;          Zhan.push (n); } else if (op[i][0]== ' Q ') {R (dian[i][0]); R (dian[i][1]); R (Dian[i][2]);          R (Rs[i]);            } else {ids[i]=zhan.top ();          Zhan.pop ();    }} root= (1+n>>1);    Buildtree ();        for (int i=1;i<=n;++i) {ma[t[i].id]=i;      for (int j=0;j<2;++j) if (t[i].ch[j]) fa[t[i].ch[j]]=i;          } for (int i=1;i<=m;++i) if (op[i][0]== ' A ') {idn=ids[i];          val=1;        Update ();          } else if (op[i][0]== ' Q ') {memcpy (qp[0],dian[i],sizeof (qp[0]));          for (int j=0;j<kd;++j) qp[1][j]=qp[0][j]+rs[i];          ans=0;          Query ();//printf ("%d\n", ans);        P (ANS), puts ("");          } else {idn=ids[i];          Val=-1;        Update (); } return 0;}

"Kd-tree" bzoj3290 Theresa and data structures