P3810 【模板】三维偏序（陌上花开）cdq分治

传送门：https://www.luogu.org/problemnew/show/P3810

cdq分治的模板题，第一层外部排序，第二层cdq归并排序，这个时候不用考虑第一次的顺序，第三次用树状数组。

注意，不要用memset，用队列保存加上的值，最后在把加上的值减去就行了。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <cstdlib>
#include   <iomanip>
#include    <bitset>
#include    <cctype>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <stack>
#include     <cmath>
#include     <queue>
#include      <list>
#include       <map>
#include       <set>
#include   <cassert>
//#include <unordered_map>
/*

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'ノ )　　Lﾉ

*/

using namespace std;
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "\n";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
//typedef __int128 bll;
typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int,pii> p3;

//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '\n'

#define boost ios::sync_with_stdio(false);cin.tie(0)
#define rep(a, b, c) for(int a = (b); a <= (c); ++ a)
#define max3(a,b,c) max(max(a,b), c);
#define min3(a,b,c) min(min(a,b), c);


const ll oo = 1ll<<17;
const ll mos = 0x7FFFFFFF;  //2147483647
const ll nmos = 0x80000000;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3f; //18
const int mod = 998244353;
const double esp = 1e-8;
const double PI=acos(-1.0);
const double PHI=0.61803399;    //黄金分割点
const double tPHI=0.38196601;


template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}

inline void cmax(int &x,int y){if(x<y)x=y;}
inline void cmax(ll &x,ll y){if(x<y)x=y;}
inline void cmin(int &x,int y){if(x>y)x=y;}
inline void cmin(ll &x,ll y){if(x>y)x=y;}
#define MODmul(a, b) ((a*b >= mod) ? ((a*b)%mod + 2*mod) : (a*b))
#define MODadd(a, b) ((a+b >= mod) ? ((a+b)%mod + 2*mod) : (a+b))

/*-----------------------showtime----------------------*/
            const int maxn = 1e5+9;
            struct node{
                int x,y,z;
                int id;
            }a[maxn],b[maxn],tmp[maxn];
            bool cmp(node a,node b){
                if(a.x != b.x) return a.x < b.x;
                if(a.y != b.y) return a.y < b.y;
                else return a.z < b.z;
            }
            int cnt[maxn],ans[maxn];

            int sum[maxn*2];
            int lowbit(int x){
                return x & (-x);
            }
            void add(int x,int c){
                while(x < maxn*2){
                    sum[x] += c;
                    x += lowbit(x);
                }
            }
            int getsum(int x){
                int res = 0;
                while(x > 0) {
                    res += sum[x];
                    x -= lowbit(x);
                }
                return res;
            }
            int lazy[maxn];
            queue<int>que;
            void cdq(int le,int ri){
                int mid = (le + ri) >> 1;
                if(ri - le <= 0) return ;
                cdq(le, mid);   cdq(mid+1, ri);

         //       memset(sum, 0, sizeof(sum));

                int p = le, q = mid+1;
                int id = 0;
                while(p <= mid && q <= ri){
                    if(a[p].y <= a[q].y) {
                        add(a[p].z, lazy[a[p].id]);
                        que.push(p);
                        tmp[++id] = a[p++];
                    }
                    else {
                        cnt[a[q].id] += getsum(a[q].z);
                        tmp[++id] = a[q++];
                    }
                }
                while(p <= mid) tmp[++id] = a[p++];
                while(q <= ri) {
                    cnt[a[q].id] += getsum(a[q].z);
                    tmp[++id] = a[q++];
                }
                while(!que.empty()){
                    int u = que.front(); que.pop();
                    add(a[u].z, -lazy[a[u].id]);
                }

                for(int i=1; i<=id; i++) a[i+le-1] = tmp[i];

            }
           // int out[maxn];
           map<p3,int>mp;


int main(){
            int n,k;
            scanf("%d%d", &n, &k);
            int tot = 0;
            rep(i, 1, n) {
                int x, y, z;
                scanf("%d%d%d", &x, &y, &z);
             //   read(x);read(y);read(z);
                p3 tp = p3(x, pii(y, z));
                if(mp.count(tp)) lazy[mp[tp]]++,cnt[mp[tp]]++;
                else {
                    tot++;
                    a[tot].id = tot;
                    a[tot].x = x;
                    a[tot].y = y;
                    a[tot].z = z;
                    mp[tp] = tot;
                    lazy[tot] = 1;
                }
            }
            sort(a+1, a+1+tot, cmp);

            cdq(1, tot);

            rep(i, 1, tot) ans[cnt[a[i].id]] += lazy[a[i].id];
            /*
            rep(i, 1, n) {
                out[a[i].id] = cnt[a[i].id];
            }
            rep(i, 1, n)
                cout<<cnt[i]<<" ";
            cout<<endl;
            */
            rep(i, 0, n-1) printf("%d\n", ans[i]);
            return 0;
}

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转载于:https://www.cnblogs.com/ckxkexing/p/10459390.html