洛谷P3810-陌上开花(三维偏序, CDQ, 树状数组)

链接:

https://www.luogu.org/problem/P3810#submit

题意:

一个元素三个属性, x, y, z, 给定求f(b) = {ax <= bx, ay <= by, az <= bz}, a的数量, 求0-(n-1)个数有多少个点满足

思路:

三维偏序, CDQ分治, 听说过, 一直想学, 先写板子题, 二维偏序就属性减到连个,搞一下.
CDQ的思想就是先对第一维排序, 左边的可以给有变贡献, 在分治下去, 同时对第二维排序, 将可用的点的第三维用数据结构维护一下, 挨个统计.

代码:

#include 
using namespace std;
const int MAXN = 2e5+10;

struct Node
{
    int x, y, z;
    int cnt;
    int ans;
}a[MAXN], b[MAXN];

int C[MAXN], Res[MAXN];
int n, k, cnt;

bool CmpX(Node a, Node b)
{
    if (a.x != b.x)
        return a.x < b.x;
    else if (a.y != b.y)
        return a.y < b.y;
    else
        return a.z < b.z;
}

bool CmpY(Node a, Node b)
{
    if (a.y != b.y)
        return a.y < b.y;
    else
        return a.z < b.z;
}

int Lowbit(int x)
{
    return x&(-x);
}

void Update(int pos, int val)
{
    while (pos <= k)
    {
        C[pos] += val;
        pos += Lowbit(pos);
    }
}

int Query(int pos)
{
    int ans = 0;
    while (pos > 0)
    {
        ans += C[pos];
        pos -= Lowbit(pos);
    }
    return ans;
}

void CDQ(int l, int r)
{
    if (l == r)//边界条件
        return ;
    int mid = (l+r)/2;
    CDQ(l, mid);
    CDQ(mid+1, r);
    sort(a+l, a+mid+1, CmpY);
    sort(a+mid+1, a+r+1, CmpY);
    int i = l, j = mid+1;
    while (j <= r)
    {
        while (i <= mid && a[j].y >= a[i].y)
            Update(a[i].z, a[i].cnt), i++;
        a[j].ans += Query(a[j].z);
        j++;
    }
    for (int z = l;z < i;z++)
        Update(a[z].z, -a[z].cnt);
}

int main()
{
    scanf("%d%d", &n, &k);
    for (int i = 1;i <= n;i++)
        scanf("%d%d%d", &b[i].x, &b[i].y, &b[i].z);
    sort(b+1, b+1+n, CmpX);
    a[1] = b[1];
    a[1].cnt = 1;
    cnt = 1;
    for (int i = 2;i <= n;i++)
    {
        if (b[i].x != a[cnt].x || b[i].y != a[cnt].y || b[i].z != a[cnt].z)
            a[++cnt] = b[i], a[cnt].cnt = 1;
        else
            a[cnt].cnt++;
    }
    CDQ(1, cnt);
    for (int i = 1;i <= cnt;i++)
        Res[a[i].ans+a[i].cnt-1] += a[i].cnt;
    for (int i = 0;i < n;i++)
        printf("%d\n", Res[i]);

    return 0;
}