HDU - 5126: stars (CDQ find points within the cube sleeve CDQ)

Meaning of the questions: is now given to the three-dimensional plane empty, a plus point operation and ask cube points.

Ideas: Consider CDQ sets of CDQ. Complexity is O (NlogN * logN * logN) , can lead to this problem.

Specifically, it is a problem of four-dimensional partial order, are four-dimensional (times, x, y, z); We know cdq can be calculated t <= T, x = X, y <= Y, in a layer on the sleeve can z <= Z a. So a cube, we split into eight points inclusion and exclusion.

The question now is then obtained (0,0,0) to the point (x, y, z) of. The first dimension T has a default sort, we first X partition. All the problems into two, and the left side of this mark o = -1, the right tag o = 1, then the normal CDQ stuffed inside the partition to be y, z for statistics. At this time, to insert a tree z array, especially to meet the needs o = -1; satisfies the query requires extra o = 1;

For CDQ: After the first dimension of a partition between two cells out after, merge sort on it, so to avoid the sort. Here can also be used for the convenience of inplace_merge, but slower than some of the handwriting.

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=400010;
struct in{
    int x,y,z,id,opt,o; //编号，正负，第一维是否在左
}s[maxn],q[maxn],fcy[maxn];
bool cmp1(in w,in v) { return w.x<v.x;}
bool cmp2(in w,in v) { return w.y<v.y;}
int b[maxn],sz,tot,cnt,ans[maxn],sum[maxn];
void add(int x,int val)
{
    for(int i=x;i<=sz;i+=(-i)&i) sum[i]+=val;
}
int query(int x)
{
    int res=0;
    for(int i=x;i;i-=(-i)&i) res+=sum[i];
    return res;
}
void cdq2(int L,int R)
{
    if(L>=R) return ;
    int Mid=(L+R)>>1,i=L,j=Mid+1,now;
    cdq2(L,Mid); cdq2(Mid+1,R);
    for(;j<=R;j++){
        for(;i<=Mid&&q[i].y<=q[j].y;i++){
            if(q[i].o==-1&&q[i].opt==0) add(q[i].z,1);
        }
        if(q[j].o==1&&q[j].id) ans[q[j].id]+=q[j].opt*query(q[j].z);
    }
    while((--i)>=L) if(q[i].o==-1&&!q[i].id) add(q[i].z,-1);

    /*i=L,j=Mid+1,now=L; 手写归并，下面cdq1同理。
    while(j<=R&&i<=Mid){
        if(q[i].y<=q[j].y) fcy[now++]=q[i++];
        else  fcy[now++]=q[j++];
    }
    while(i<=Mid) fcy[now++]=q[i++];
    while(j<=R)  fcy[now++]=q[j++];
    rep(i,L,R) q[i]=fcy[i];*/
    inplace_merge(q+L,q+Mid+1,q+R+1,cmp2); //归并排序
}
void cdq1(int L,int R)
{
    if(L>=R) return ;
    int Mid=(L+R)>>1,i=L,j=Mid+1,now=L;
    cdq1(L,Mid);  cdq1(Mid+1,R);
    //sort(s+L,s+Mid+1,cmp1); sort(s+Mid+1,s+R+1,cmp1);
    tot=0;
    while(j<=R&&i<=Mid){
        if(s[i].x<=s[j].x){
            fcy[now++]=s[i];
            q[++tot]=s[i++],q[tot].o=-1;
            if(s[i-1].id) tot--;
        }
        else {
            fcy[now++]=s[j];
            q[++tot]=s[j++],q[tot].o=1;
            if(!s[j-1].id) tot--;
        }
    }
    while(i<=Mid) {
        fcy[now++]=s[i];
        q[++tot]=s[i++],q[tot].o=-1;
        if(s[i-1].id) tot--;
    }
    while(j<=R) {
        fcy[now++]=s[j];
        q[++tot]=s[j++],q[tot].o=1;
        if(!s[j-1].id) tot--;
    }
    rep(i,L,R) s[i]=fcy[i];
    //inplace_merge(s+L,s+Mid+1,s+R+1,cmp1);
    cdq2(1,tot);
}
void solve()
{
    int N,Q=0,opt,x1,y1,z1,x2,y2,z2;
    sz=0; tot=0;
    scanf("%d",&N);
    rep(i,1,N){
        scanf("%d",&opt);
        if(opt&1){
            scanf("%d%d%d",&x1,&y1,&z1);
            b[++sz]=z1;
            s[++tot]=in{x1,y1,z1,0,0,0};
        }
        else {
            scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
            b[++sz]=z1; b[++sz]=z2; b[++sz]=z1-1; ans[++Q]=0;
            s[++tot]=in{x2,y2,z2,Q,1,0};
            s[++tot]=in{x1-1,y1-1,z1-1,Q,-1,0};
            s[++tot]=in{x1-1,y2,z2,Q,-1,0};
            s[++tot]=in{x1-1,y1-1,z2,Q,1,0};
            s[++tot]=in{x1-1,y2,z1-1,Q,1,0};
            s[++tot]=in{x2,y1-1,z2,Q,-1,0};
            s[++tot]=in{x2,y2,z1-1,Q,-1,0};
            s[++tot]=in{x2,y1-1,z1-1,Q,1,0};
        }
    }
    sort(b+1,b+sz+1);
    cnt=unique(b+1,b+sz+1)-(b+1);
    rep(i,1,tot) s[i].z=lower_bound(b+1,b+sz+1,s[i].z)-b;
    cdq1(1,tot);
    rep(i,1,Q) printf("%d\n",ans[i]);
}
int main()
{
    int T; scanf("%d",&T);
    while(T--) solve();
    return 0;
}