HDU - 5618 【CDQ分治+樹狀數組求解三維偏序】

題目鏈接

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  T組測試，有N個點然後給出每個點的三維座標，然後求每一位座標都小於等於它的點的數量有多少個。

  簡單的三維偏序問題，但是需要注意的是，完全相等的情況，所以一開始的時候預處理一下即可。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <unordered_map>
#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define Big_INF 0x3f3f3f3f3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
//const int maxN = 6;
int N, _UP, rid[maxN];
struct node
{
    int x, y, z, id, cnt;
    inline void In(int ith) { scanf("%d%d%d", &x, &y, &z); id = ith; }
    friend bool operator < (node e1, node e2) { return e1.x == e2.x ? (e1.y == e2.y ? e1.z < e2.z : e1.y < e2.y) : e1.x < e2.x; }
    friend bool operator == (node e1, node e2) { return e1.x == e2.x && e1.y == e2.y && e1.z == e2.z; }
} t[maxN], cop[maxN], q[maxN];
ll ans[maxN], tree[maxN];
void update(int x, ll val) { while(x <= 100000) { tree[x] += val; x += lowbit(x); } }
ll query(int x) { ll sum = 0; while(x) { sum += tree[x]; x -= lowbit(x); } return sum; }
void cdq(int l, int r)
{
    if(l == r) return;
    int mid = HalF;
    cdq(l, mid); cdq(mid + 1, r);
    int ql = l, qr = mid + 1, pos = l;
    while(ql <= mid || qr <= r)
    {
        if(qr > r || (ql <= mid && q[ql].y <= q[qr].y))
        {
            update(q[ql].z, q[ql].cnt);
            cop[pos++] = q[ql++];
        }
        else
        {
            ans[q[qr].id] += query(q[qr].z);
            cop[pos++] = q[qr++];
        }
    }
    for(int i=l; i<=mid; i++) update(q[i].z, -q[i].cnt);
    for(int i=l; i<=r; i++) q[i] = cop[i];
}
int main()
{
    int T; scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &N);
        for(int i=1; i<=N; i++) ans[i] = 0;
        for(int i=1; i<=N; i++) t[i].In(i);
        sort(t + 1, t + N + 1);
        q[_UP = 1] = t[1];
        rid[t[1].id] = 1;
        q[_UP].id = 1; q[1].cnt = 1;
        for(int i=2; i<=N; i++)
        {
            if(t[i] == q[_UP]) { q[_UP].cnt++; rid[t[i].id] = _UP; }
            else
            {
                q[++_UP] = t[i];
                rid[t[i].id] = _UP;
                q[_UP].id = _UP;
                q[_UP].cnt = 1;
            }
        }
        cdq(1, _UP);
        for(int i=1; i<=_UP; i++) ans[q[i].id] += q[i].cnt - 1;
        for(int i=1; i<=N; i++) printf("%lld\n", ans[rid[i]]);
    }
    return 0;
}