hdu 5126 stars (四维偏序，离线，CDQ套CDQ套树状数组）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5126

思路：支持离线，那么我们可以用两次CDQ分治使四维降为二维，降成二维后排个序用树状数组维护下就好了

实现代码：

#include<bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int M = 5e5+10;

int ans[M],c[M];
struct node
{
    int x,y,z;
    int kind,id;
    node(){}
    node(int x1,int y1,int z1,int k,int i):x(x1),y(y1),z(z1),kind(k),id(i){}
};
vector<node>q,q1,q2;
vector<int>v;
void init(){
    q.clear();
    v.clear();
    memset(c,0,sizeof(c));
}

bool cmp(node a,node b){
    if(a.x == b.x) return a.id < b.id;
    return a.x < b.x;
}

bool cmp1(node a,node b){
    if(a.y == b.y) return a.id < b.id;
    return a.y < b.y;
}

int lowbit(int x){
    return x&-x;
}

void add(int x,int val){
    while(x <= v.size()){
        c[x] += val;
        x += lowbit(x);
    }
}

int getsum(int x){
    int sum = 0;
    while(x > 0){
        sum += c[x];
        x -= lowbit(x);
    }
    return sum;
}

void countstar(){
    for(int i = 0;i < q2.size();i ++){
        if(q2[i].kind == 0) add(q2[i].z,1);
        else ans[q2[i].id] += q2[i].kind*getsum(q2[i].z);
       // cout<<q2[i].id<<" "<<ans[q2[i].id]<<endl;
    }
    for(int i = 0;i < q2.size();i ++)
        if(q2[i].kind == 0) add(q2[i].z,-1);
}

void cdq1(int l,int r){
    if(l >= r) return ;
    int mid = (l + r) >> 1;
    cdq1(l,mid); cdq1(mid+1,r);
    q2.clear();
    for(int i = l;i <= mid;i ++)
        if(q1[i].kind == 0) q2.push_back(q1[i]);
    for(int i = mid+1;i <= r;i ++)
        if(q1[i].kind) q2.push_back(q1[i]);
    sort(q2.begin(),q2.end(),cmp1);
    countstar();
}

void cdq(int l,int r){
    if(l >= r) return ;
    int mid = (l + r) >> 1;
    cdq(l,mid); cdq(mid+1,r);
    q1.clear();
    for(int i = l;i <= mid;i ++)
        if(q[i].kind == 0) q1.push_back(q[i]);
    for(int i = mid+1;i <= r;i ++)
        if(q[i].kind) q1.push_back(q[i]);
    sort(q1.begin(),q1.end(),cmp);
    cdq1(0,q1.size()-1);
}


int main()
{
    int t,n,op,x,y,z,x1,y1,z1,x2,y2,z2;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        init();
        for(int i = 0;i < n;i ++){
            scanf("%d",&op);
            if(op == 1){
                scanf("%d%d%d",&x,&y,&z);
                q.push_back(node(x,y,z,0,i));
                v.push_back(z);
                ans[i] = -1;
            }
            else{
                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
                x1--; y1--; z1--;
                q.push_back(node(x1,y1,z1,-1,i));
                q.push_back(node(x2,y2,z2,1,i));
                q.push_back(node(x1,y2,z2,-1,i));
                q.push_back(node(x2,y1,z2,-1,i));
                q.push_back(node(x2,y2,z1,-1,i));
                q.push_back(node(x2,y1,z1,1,i));
                q.push_back(node(x1,y2,z1,1,i));
                q.push_back(node(x1,y1,z2,1,i));
                v.push_back(z1); v.push_back(z2);
                ans[i] = 0;
            }

        }
        //离散化
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        for(int i = 0;i < q.size();i ++)
            q[i].z = (lower_bound(v.begin(),v.end(),q[i].z)-v.begin())+1;
        cdq(0,q.size()-1);
        for(int i = 0;i < n;i ++)
            if(ans[i] != -1) printf("%d
",ans[i]);
    }
    return 0;
}