cdq分治解决三维偏序

时间：2023-03-09 03:36:34
cdq分治解决三维偏序

问题背景

在三维坐标系中有n个点，坐标为(xi,yi,zi). 定义一个点A比一个点B小，当且仅当xA<=xB,yA<=yB,zA<=zB。问对于每个点，有多少个点比它小。(n<=1e5)

其实就是离散数学里的偏序的概念啦，只不过是到了三维。回顾一下偏序的概念：

偏序关系：自反，反对称且传递，符号<

然后先考虑一下二维偏序吧。可以用最长上升子序列LIS来做，然后我们今天讨论一种特殊分治的做法，这种算法是由08年集训队的陈丹琦提出来的，因此叫cdq分治。

主要思想就是先按照第一维排序。然后遍历每一个点，此时我们要统计的就是前面的点中比这个点的y坐标要小的点的个数。我们用一个树状数组来维护y坐标这个信息，于是就只要得到getsum(y)就行了。

三维的CDQ分治呢，做法如下：

第一维排序，第二维CDQ分治，第三维树状数组。

第一维比如先按照x坐标排序。在第二维的CDQ分治时，我们对每一个自区间，先按照y排序，计算左边对右边的影响的时候:

  1. 左边的x都小于右边

  2. 在每一边y也是依次递增的

  3. 我们只要扫描右边，把左边y小于等于当前的y坐标的z坐标更新到树状数组，统计目前树状数组z坐标小于自己的就是偏序<的点的个数。

复杂度分析

根据主定理：

T(n)=2T(n2)+O(kn)=O(knlogn)T(n)=2T(n2)+O(kn)=O(knlogn)

T(n)=2T(n2)+O(knlogn)=O(knlog2 n)

例题

BZOJ 3262 陌上花开

牛客网的一套题

后面这个题虽然是个裸三维偏序，不过也可以转化成三个二维偏序。

另附一个BZOJ3262的别人的代码，可做模板。

//bzoj 3262

//1维排序，二维分治，3维树状数组

#include <bits/stdc++.h>

using namespace std;

const int maxn = 200005;

int n, m, ans[maxn], tree[maxn*4];

struct node{

    int a, b, c, s, ans; //s处理相同连续,ans比当前美丽值小个数

    node(){}

    node(int a, int b, int c, int s, int ans) : a(a), b(b), c(c), s(s), ans(ans) {}

    bool operator < (const node &rhs) const{ //按y排序

        if(b == rhs.b) return c < rhs.c;

        return b < rhs.b;

    }

}a[maxn], p[maxn];

bool cmp(node x, node y){ //按照x排序

    if(x.a == y.a && x.b == y.b) return x.c < y.c;

    if(x.a == y.a) return x.b < y.b;

    return x.a < y.a;

}

namespace BIT{

    inline int lowbit(int x) {return x&-x;}

    inline void update(int x, int y){for(int i = x; i <= m; i+=lowbit(i)) tree[i] += y;}

    inline int query(int x){int res = 0; for(int i = x; i; i -= lowbit(i)) res += tree[i]; return res;}

}

using namespace BIT;

void CDQ(int l, int r)

{

    if(l == r) return;

    int mid = (l + r) >> 1;

    CDQ(l, mid);

    CDQ(mid+1, r);

    sort(p + l, p + mid + 1);

    sort(p + mid + 1, p + r + 1);

    int i = l, j = mid + 1;

    while(j <= r){

        while(i <= mid && p[i].b <= p[j].b){

            update(p[i].c, p[i].s);

            i++;

        }

        p[j].ans += query(p[j].c);

        j++;

    }

    for(int j = l; j < i; j++) update(p[j].c, -p[j].s);

}

int main(){

    int nn;

    scanf("%d%d", &nn, &m);

    for(int i = 1; i <= nn; i++){

        scanf("%d%d%d", &a[i].a, &a[i].b, &a[i].c);

    }

    sort(a + 1, a + nn + 1, cmp); //按照x排

    int cnt = 0; //unique

    for(int i = 1; i <= nn; i++){

        cnt++;

        if(a[i].a != a[i+1].a || a[i].b != a[i+1].b || a[i].c != a[i+1].c){

            p[++n] = a[i];

            p[n].s = cnt;

            cnt = 0;

        }

    }

    CDQ(1, n);

    for(int i = 1; i <= n; i++){

        ans[p[i].ans + p[i].s - 1] += p[i].s;

    }

    for(int i = 0; i < nn; i++){

        printf("%d\n", ans[i]);

    }

    return 0;

}

总结

从二维偏序出发，了解了三维偏序的CDQ分治做法。

在这类问题中通常将时间(操作序列)作为第一维，剩下的二维问题使用CDQ分治和数据结构。

这种问题也可以用树套树做，据说很烦，树状数组套个什么Treap啥的，总之比这个CDQ要难写很多。

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