CDQ分治笔记（持续更新）

最新推荐文章于 2024-10-20 06:30:00 发布

qq_45843001

最新推荐文章于 2024-10-20 06:30:00 发布

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分类专栏：进阶课文章标签：算法分治算法

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CDQ分治

即对于每个i,有多少个j使得 $a_j<=a_i$ , $b_j<=b_i$ , $c_j<=c_i$

一维：

排序即可

二维：

①：树状数组解决即可
值域统计后插入元素即可
②：分治
加双关键字排序（便于处理 $a_j<=a_i,b_j<=b_i$ ）的情况
$e g$ 逆序对：
①：i,j都在左半边
②：i,j都在右半边
③:j在左，i在右
①和②递归即可，至于③，考虑满足第二维的限制即可，使用双指针算法，时间复杂度 $O (n l o g n)$

三维

①：三关键字排序
②：将满足i，j要求的数对分成三大类
A:i,j均在左半边（递归求）
B:i,j均在右半边（递归求）
C:j在左，i在右
按B排序
$对于每个i, 通过双指针找到bi<=bj的位置，则在1至j的所有数里找到满足<=c_i的数的个数（树状数组）$

时间复杂度 $O(nlog^2n)$

备注：对于 $a_i==a_j,b_i == b_j,c_i == c_j$ 的情况，去除掉，并记录个数,最后在答案加上（个数-1）

模板题
代码+详细注释

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100010, M = 200010;

int n, m;
struct Data
{
    int a, b, c, s, res;

    bool operator< (const Data& t) const
    {
        if (a != t.a) return a < t.a;
        if (b != t.b) return b < t.b;
        return c < t.c;
    }
    bool operator== (const Data& t) const
    {
        return a == t.a && b == t.b && c == t.c;
    }
}q[N], w[N];
int tr[M], ans[N];

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

void add(int x, int v)
{
    for (int i = x; i < M; i += lowbit(i)) tr[i] += v;
}

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

void merge_sort(int l, int r)
{
    if (l >= r) return;
    int mid = l + r >> 1;
    merge_sort(l, mid), merge_sort(mid + 1, r);
    int i = l, j = mid + 1, k = 0;
    while (i <= mid && j <= r)
        if (q[i].b <= q[j].b) add(q[i].c, q[i].s), w[k ++ ] = q[i ++ ];
        else q[j].res += query(q[j].c), w[k ++ ] = q[j ++ ];
    while (i <= mid) add(q[i].c, q[i].s), w[k ++ ] = q[i ++ ];
    while (j <= r) q[j].res += query(q[j].c), w[k ++ ] = q[j ++ ];
    for (i = l; i <= mid; i ++ ) add(q[i].c, -q[i].s);
    for (i = l, j = 0; j < k; i ++, j ++ ) q[i] = w[j];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        q[i] = {a, b, c, 1};
    }
    sort(q, q + n);

    int k = 1;
    for (int i = 1; i < n; i ++ )
        if (q[i] == q[k - 1]) q[k - 1].s ++ ;
        else q[k ++ ] = q[i];

    merge_sort(0, k - 1);
    for (int i = 0; i < k; i ++ )
        ans[q[i].res + q[i].s - 1] += q[i].s;

    for (int i = 0; i < n; i ++ ) printf("%d\n", ans[i]);

    return 0;
}

应用

一：老C的任务
思路：考虑转化为CDQ分治
这对于每个点，求出其作为左上顶点的矩形中所包含的点的权值之和。（即二维前缀和）
而zi只有1，0两种取值，不需使用树状数组，使用一个变量维护即可。
代码如下：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 500010;

int n, m;
struct Data
{
    int x, y, z, p, id, sign;
    LL sum;

    bool operator< (const Data& t) const
    {
        if (x != t.x) return x < t.x;
        if (y != t.y) return y < t.y;
        return z < t.z;
    }
}q[N], w[N];
LL ans[N];

void merge_sort(int l, int r)
{
    if (l >= r) return;
    int mid = l + r >> 1;
    merge_sort(l, mid), merge_sort(mid + 1, r);
    int i = l, j = mid + 1, k = 0;
    LL sum = 0;
    while (i <= mid && j <= r)
        if (q[i].y <= q[j].y) sum += !q[i].z * q[i].p, w[k ++ ] = q[i ++ ];
        else q[j].sum += sum, w[k ++ ] = q[j ++ ];
    while (i <= mid) sum += !q[i].z * q[i].p, w[k ++ ] = q[i ++ ];
    while (j <= r) q[j].sum += sum, w[k ++ ] = q[j ++ ];
    for (i = l, j = 0; j < k; i ++, j ++ ) q[i] = w[j];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ )
    {
        int x, y, p;
        scanf("%d%d%d", &x, &y, &p);
        q[i] = {x, y, 0, p};
    }
    int k = n;
    for (int i = 1; i <= m; i ++ )
    {
        int x1, y1, x2, y2;
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        q[k ++ ] = {x2, y2, 1, 0, i, 1};
        q[k ++ ] = {x1 - 1, y2, 1, 0, i, -1};
        q[k ++ ] = {x2, y1 - 1, 1, 0, i, -1};
        q[k ++ ] = {x1 - 1, y1 - 1, 1, 0, i, 1};
    }

    sort(q, q + k);
    merge_sort(0, k - 1);

    for (int i = 0; i < k; i ++ )
        if (q[i].z)
            ans[q[i].id] += q[i].sum * q[i].sign;

    for (int i = 1; i <= m; i ++ ) printf("%lld\n", ans[i]);
    return 0;
}

二：动态逆序对

思路：现在已经有两维，第三维为时间戳，即当前该数被删除的时间
需要记录

代码如下：

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int N = 100010;

int n, m;
struct Data
{
    int a, t, res;
}q[N], w[N];
int tr[N], pos[N];
LL ans[N];

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

void add(int x, int v)
{
    for (int i = x; i < N; i += lowbit(i)) tr[i] += v;
}

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

void merge_sort(int l, int r)
{
    if (l >= r) return;
    int mid = l + r >> 1;
    merge_sort(l, mid), merge_sort(mid + 1, r);
    int i = mid, j = r;
    while (i >= l && j > mid)
        if (q[i].a > q[j].a) add(q[i].t, 1), i -- ;
        else q[j].res += query(q[j].t - 1), j -- ;
    while (j > mid) q[j].res += query(q[j].t - 1), j -- ;
    for (int k = i + 1; k <= mid; k ++ ) add(q[k].t, -1);

    j = l, i = mid + 1;
    while (j <= mid && i <= r)
        if (q[i].a < q[j].a) add(q[i].t, 1), i ++ ;
        else q[j].res += query(q[j].t - 1), j ++ ;
    while (j <= mid) q[j].res += query(q[j].t - 1), j ++ ;
    for (int k = mid + 1; k < i; k ++ ) add(q[k].t, -1);

    i = l, j = mid + 1;
    int k = 0;
    while (i <= mid && j <= r)
        if (q[i].a <= q[j].a) w[k ++ ] = q[i ++ ];
        else w[k ++ ] = q[j ++ ];
    while (i <= mid) w[k ++ ] = q[i ++ ];
    while (j <= r) w[k ++ ] = q[j ++ ];

    for (i = l, j = 0; j < k; i ++, j ++ ) q[i] = w[j];
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i ++ )
    {
        scanf("%d", &q[i].a);
        pos[q[i].a] = i;
    }
    for (int i = 0, j = n; i < m; i ++ )
    {
        int a;
        scanf("%d", &a);
        q[pos[a]].t = j -- ;
        pos[a] = -1;
    }

    for (int i = 1, j = n - m; i <= n; i ++ )
        if (pos[i] != -1)
            q[pos[i]].t = j -- ;

    merge_sort(0, n - 1);

    for (int i = 0; i < n; i ++ ) ans[q[i].t] = q[i].res;
    for (int i = 2; i <= n; i ++ ) ans[i] += ans[i - 1];
    for (int i = 0, j = n; i < m; i ++, j -- ) printf("%lld\n", ans[j]);

    return 0;
}