洛谷P3810（cdq）

題目鏈接：https://www.luogu.org/problemnew/show/P3810

思路：cdq模板題，三維分別用排序、樹狀數組、cdq處理

#pragma GCC optimize(2)
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%lld",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(1000)
#define pb push_back
#define eps 1e-8
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n,k,ans[100008],b[200008];
struct as{
int a,b,c,ans,num;}a[100008],c[100008];
bool cmp(as a,as b)
{
    if(a.a==b.a)
    {
        if(a.b==b.b) return a.c<b.c;
        return a.b<b.b;
    }
    return a.a<b.a;
}
bool cmp1(as a,as b)
{
    if(a.b==b.b) return a.c<b.c;
    return a.b<b.b;
}
void add(int x,int y)  {while(x<=k) b[x]+=y,x+=x&-x;}
int sum(int x)
{
    int res=0;
    while(x>0) res+=b[x],x-=x&-x;
    return res;
}
void cdq(int l,int r)
{
    if(l==r) return;
    int mid=(l+r)>>1;
    cdq(l,mid),cdq(mid+1,r);
    sort(c+l,c+mid+1,cmp1);
    sort(c+mid+1,c+r+1,cmp1);
    int zz=l;
    FOR(i,mid+1,r)
    {
        while(c[zz].b<=c[i].b&&zz<=mid) add(c[zz].c,c[zz].num),zz++;
        c[i].ans+=sum(c[i].c);
        //cout<<c[i].ans<<endl;
    }
    FOR(i,l,zz-1) add(c[i].c,-c[i].num);
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>n>>k;
    FOR(i,1,n) si(a[i].a),si(a[i].b),si(a[i].c);
    sort(a+1,a+n+1,cmp);
    int qw=0,zz=0;
    FOR(i,1,n)
    {
        zz++;
        if((a[i].a^a[i+1].a)||(a[i].b^a[i+1].b)||(a[i].c^a[i+1].c)) c[++qw]=a[i],c[qw].num=zz,zz=0;
    }
    cdq(1,qw);
   // FOR(i,1,qw) cout<<sum(i)<<endl;
    FOR(i,1,qw) ans[c[i].ans+c[i].num-1]+=c[i].num;
    FOR(i,0,n-1) printf("%d\n",ans[i]);
    return 0;
}