bzoj3262: 陌上花开 三维偏序cdq分治

三维偏序裸题，cdq分治时，左侧的x一定比右侧x小，然后分别按y排序，对于左侧元素按y大小把z依次插入到树状数组里，其中维护每个左侧元素对右侧元素的贡献，在bit查询即可

/**************************************************************
    Problem: 3262
    User: walfy
    Language: C++
    Result: Accepted
    Time:3844 ms
    Memory:28640 kb
****************************************************************/
 
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define ld long double
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pll pair<ll,ll>
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define Max(a,  b) ((a)>(b)?(a):(b))
#define Min(a,  b) ((a)<(b)?(a):(b))
#define fio ios::sync_with_stdio(false);cin.tie(0)
inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}
 
using namespace std;
 
const double eps=1e-8;
const ll INF=0x3f3f3f3f3f3f3f3f;
const int N=1000000+10,maxn=50000+10,inf=0x3f3f3f3f;
 
struct node{
    int x,y,z,w,ans;
    bool operator <(const node &rhs)const{
        return x<rhs.x || x==rhs.x&&y<rhs.y ||x==rhs.x&&y==rhs.y&&z<rhs.z;
    }
}a[N];
inline bool cmp(node a,node b){return a.y<b.y||a.y==b.y&&a.z<b.z||a.y==b.y&&a.z==b.z&&a.x<b.x;}
struct BIT{
    int sum[N];
    void add(int i,int v)
    {
        for(;i<N;i+=i&(-i))sum[i]+=v;
    }
    int query(int i)
    {
        int ans=0;
        for(;i;i-=i&(-i))ans+=sum[i];
        return ans;
    }
}b;
int num[N];
void cdq(int l,int r)
{
    if(l==r){a[l].ans+=a[l].w-1;return ;}
    int m=(l+r)>>1;
    cdq(l,m);cdq(m+1,r);
    sort(a+l,a+m+1,cmp);sort(a+m+1,a+r+1,cmp);
    int now=l;
    for(int i=m+1;i<=r;i++)
    {
        while(now<=m&&a[now].y<=a[i].y)b.add(a[now].z,a[now].w),now++;
        a[i].ans+=b.query(a[i].z);
    }
    for(int i=l;i<now;i++)
        b.add(a[i].z,-a[i].w);
}
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
        scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z),a[i].ans=1;
    sort(a+1,a+1+n);
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(i!=1&&a[i].x==a[i-1].x&&a[i].y==a[i-1].y&&a[i].z==a[i-1].z)a[cnt].w++;
        else a[++cnt]=a[i],a[cnt].w=1;
    }
    cdq(1,cnt);
    sort(a+1,a+1+cnt);
    for(int i=1;i<=cnt;i++)num[a[i].ans]+=a[i].w;
    for(int i=1;i<=n;i++)printf("%d\n",num[i]);
    return 0;
}
/********************
10 3
3 3 3
2 3 3
2 3 1
3 1 1
3 1 2
1 3 1
1 1 2
1 2 2
1 3 2
1 2 1
********************/