HDU 5126 stars（求四维偏序，cdq分治+树状数组）

题目链接：http://hdu.hustoj.com/showproblem.php?pid=5126

Problem Description

John loves to see the sky. A day has Q times. Each time John will find a new star in the sky, or he wants to know how many stars between (x1,y1,z1) and (x2,y2,z2).

Input

The first line contains a single integer T(1≤T≤10) (the data for Q>100 less than 6 cases),, indicating the number of test cases.
The first line contains an integer Q(1≤Q≤50000),indicating how many times in a day.
Next Q lines contain some integers, first input an integer A(1≤A≤2).If A=1 then input 3 integers x, y and z, indicating a coordinate of one star.. If A=2 then input 6 integers x1,y1,z1,x2,y2,z2(1≤x,y,z,x1,y1,z1,x2,y2,z2≤109,x1≤x2,y1≤y2,z1≤z2).

Output

For each “A=2”,output an integer means how many stars in such a section.

Sample Input

1 1 1 1

2 1 1 1 1 1 1

1 2 2 2

1 1 1 2

2 1 1 1 2 2 2

1 3 3 3

1 4 4 4

1 5 5 5

1 6 6 6

2 1 1 1 6 6 6

2 3 3 3 6 6 6

1 1 1 1

2 1 1 1 1 1 1

1 2 2 2

1 1 1 2

2 1 1 1 2 2 2

1 3 3 3

1 4 4 4

1 5 5 5

1 6 6 6

2 1 1 1 6 6 6

2 3 3 3 6 6 6

Sample Output

题目大意：有两种操作，一种是插入(x,y,z)这个坐标,第二种是查询(x1,y1,z1)到(x2,y2,z2)(x1<=x2,y1<=y2,z1<=z2)的长方体包含多少个点。

题目思路：拆分成8个点容斥，求四维偏序。注意，需离散化z，不然树状数组存不下。

可先看BZOJ 1176这个题，求二维的子矩阵和，然后再来求这个会轻松很多，题目链接：https://blog.csdn.net/baodream/article/details/82683690

AC代码：

#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<map>

using namespace std;

#define FOU(i,x,y) for(int i=x;i<=y;i++)
#define FOD(i,x,y) for(int i=x;i>=y;i--)
#define MEM(a,val) memset(a,val,sizeof(a))
#define PI acos(-1.0)

const double EXP = 1e-9;
typedef long long ll;
typedef unsigned long long ull;
const int INF = 0x3f3f3f3f;
const ll MINF = 0x3f3f3f3f3f3f3f3f;
const double DINF = 0xffffffffffff;
const int mod = 1e9+7;
const int N = 1e6+5;

struct node{
    int typ;
    int x,y,z;
    int id;
    int val;
    bool flag;
    int pos;
}a[N],tmp[N],tmp1[N];

int vec[N],m,tot,ans[N];

void Hash(){
    sort(vec+1,vec+1+m);
    int len = unique(vec+1,vec+1+m)-vec-1;
    for(int i=1;i<=m;i++)
        a[i].z = lower_bound(vec+1,vec+len+1,a[i].z)-vec;
}

const int maxn = 500005;
int tree[maxn];
int lowbit(int x){
    return x&(-x);
}

void Add(int x,int val){
    while(x<maxn){
        tree[x]+=val;
        x+=lowbit(x);
    }
}
int Query(int x){
    int res=0;
    while(x){
        res+=tree[x];
        x-=lowbit(x);
    }
    return res;
}

void clearr(int x){
    while(x<maxn){
        if(tree[x]==0)
            break;
        tree[x]=0;
        x+=lowbit(x);
    }
}

void CDQ2(int l,int r){
    if(l>=r) return ;
    int mid = l+r>>1;
    CDQ2(l,mid);
    CDQ2(mid+1,r);
    //printf("l=%d,r=%d\n",l,r);
    int p=l,q=mid+1,k=l;
    while(p<=mid&&q<=r){
        if(tmp[p].y<=tmp[q].y){
            if(tmp[p].flag&&tmp[p].typ==1)
                Add(tmp[p].z,1);
            tmp1[k++] = tmp[p++];
        }
        else{
            if(tmp[q].flag==0&&tmp[q].typ==2){
                ans[tmp[q].pos]+=Query(tmp[q].z)*tmp[q].val;
                //printf("-------1:%d %d %d %d %d\n",tmp[q].id,tmp[q].z,tmp[q].val,Query(tmp[q].z),Query(tmp[q].z)*tmp[q].val);
            }
            tmp1[k++] = tmp[q++];
        }
    }
    while(p<=mid){
        tmp1[k++] = tmp[p++];
    }
    while(q<=r){
        if(tmp[q].flag==0&&tmp[q].typ==2){
            ans[tmp[q].pos]+=Query(tmp[q].z)*tmp[q].val;
            //printf("-------2:%d %d %d %d %d\n",tmp[q].id,tmp[q].z,tmp[q].val,Query(tmp[q].z),Query(tmp[q].z)*tmp[q].val);
        }
        tmp1[k++] = tmp[q++];
    }
    for(int i=l;i<=r;i++) clearr(tmp[i].z),tmp[i]=tmp1[i];
}

void CDQ(int l,int r){
    if(l>=r) return;
    int mid = l+r>>1;
    CDQ(l,mid);
    CDQ(mid+1,r);
    int p=l,q=mid+1,k=l;
    while(p<=mid&&q<=r){
        if(a[p].x<=a[q].x){
            tmp[k++] = a[p++];
            tmp[k-1].flag = 1;
        }
        else{
            tmp[k++] = a[q++];
            tmp[k-1].flag = 0;
        }
    }
    while(p<=mid){
        tmp[k++] = a[p++];
        tmp[k-1].flag = 1;
    }
    while(q<=r){
        tmp[k++] = a[q++];
        tmp[k-1].flag = 0;
    }
    for(int i=l;i<=r;i++) a[i] = tmp[i];
    CDQ2(l,r);
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    std::ios::sync_with_stdio(false);
    int T,q,type;
    scanf("%d",&T);
    while(T--){
        scanf("%d",&q);
        m=0,tot=0;
        int x1,y1,z1,x2,y2,z2;
        while(q--){
            scanf("%d",&type);
            if(type==1){
                scanf("%d%d%d",&x1,&y1,&z1);
                a[++m] = node{1,x1,y1,z1,m,1,0,0};
                vec[m] = z1;
            }
            else{
                scanf("%d%d%d%d%d%d",&x1,&y1,&z1,&x2,&y2,&z2);
                ++tot;
                a[++m] = node{2,x2,y2,z2,m,1,0,tot};        vec[m] = z2;
                a[++m] = node{2,x1-1,y2,z2,m,-1,0,tot};     vec[m] = z2;
                a[++m] = node{2,x2,y1-1,z2,m,-1,0,tot};     vec[m] = z2;
                a[++m] = node{2,x2,y2,z1-1,m,-1,0,tot};     vec[m] = z1-1;
                a[++m] = node{2,x2,y1-1,z1-1,m,1,0,tot};    vec[m] = z1-1;
                a[++m] = node{2,x1-1,y2,z1-1,m,1,0,tot};    vec[m] = z1-1;
                a[++m] = node{2,x1-1,y1-1,z2,m,1,0,tot};    vec[m] = z2;
                a[++m] = node{2,x1-1,y1-1,z1-1,m,-1,0,tot}; vec[m] = z1-1;
            }
        }
        Hash();
        MEM(ans,0);
        MEM(tree,0);
        CDQ(1,m);
        for(int i=1;i<=tot;i++)
            printf("%d\n",ans[i]);
    }
    return 0;
}

HDU 5126 stars（求四维偏序，cdq分治+树状数组）

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