初次接觸CDQ分治，感覺真的挺厲害的。

整體思路即分而治之，再用之前處理出來的答案統計之後的答案。

大概流程是：

對於區間 l ~ r :

1.處理 l ~mid, mid + 1 ~ r 的答案

2.分別排序規整

3.計算 l ~ mid 中每一個數對 mid + 1 ~ r 中的答案的貢獻， 累加

4.得到區間l ~ r的答案

CDQ分治我一共也才做了兩道題目， 就一起整理在這裏了。大體都差不多，CDQ+樹狀數組分別維護兩個維度。

1.三維偏序

#include <bits/stdc++.h>
using namespace std;
#define maxn 3000000
#define
 
 lowbit(x) x &(-x)
int n, k, tot, ans[maxn], c[maxn];
struct node
{
    int a, b, c, ans, cnt;
}num[maxn], a[maxn];

int read()
{
    int x = 0, k = 1;
    char c;
    c = getchar();
    while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘
 
9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * k;
}

bool cmp(node a, node b)
{
    if(a.a != b.a) return a.a < b.a;
    if(a.b != b.b) return a.b < b.b;
    return a.c < b.c;
}

bool cmp2(node a, node b)
{
    if(a.b != b.b) return a.b < b.b;
    return a.c < b.c; 
}

void Update(int
 
 x, int v)
{
    for(int i = x; i <= k; i += lowbit(i))
        c[i] += v;
}

int Query(int x)
{
    int ans = 0;
    for(int i = x; i; i -= lowbit(i))
        ans += c[i];
    return ans;
}

void cdq(int l, int r)
{
    int mid = (l + r) >> 1;
    if(r - l >= 2) cdq(l, mid), cdq(mid + 1, r);
    if(r == l) return;
    sort(num + l, num + mid + 1, cmp2);
    sort(num + mid + 1, num + r + 1, cmp2);
    int i = l, j = mid + 1;
    while(i <= mid && j <= r)
    {
        if(num[i].b <= num[j].b) Update(num[i].c, num[i].cnt), i ++;
        else num[j].ans += Query(num[j].c), j ++;
    }
    while(i <= mid) Update(num[i].c, num[i].cnt), i ++;
    while(j <= r) num[j].ans += Query(num[j].c), j ++;
    for(int i = l; i <= mid; i ++) Update(num[i].c, -num[i].cnt);
}

int main()
{
    n = read(), k = read();
    for(int i = 1; i <= n; i ++)
        a[i].a = read(), a[i].b = read(), a[i].c = read();
    sort(a + 1, a + 1 + n, cmp);
    for(int i = 1; i <= n;)
    {
        int j = 1;
        while(i + j <= n && a[i].a == a[i + j].a && a[i].b == a[i + j].b && a[i].c == a[i + j].c) j ++;
        num[++ tot] = a[i];
        num[tot].cnt = j;
        i += j;
    }
    cdq(1, tot);
    for(int i = 1; i <= tot; i ++) ans[num[i].ans + num[i].cnt - 1] += num[i].cnt;
    for(int i = 0; i < n; i ++) printf("%d\n", ans[i]);
    return 0;
}

2.動態逆序對

#include <bits/stdc++.h>
using namespace std;
#define maxn 2000000
#define ll long long
#define lowbit(x) x & (-x)
int n, m, timer, a[maxn], b[maxn], d[maxn], t[maxn];
ll ans[maxn], c[maxn];
struct node
{
    int t, num, pl;
    ll ans;
}w[maxn];

int read()
{
    int x = 0, k = 1;
    char c;
    c = getchar();
    while(c < ‘0‘ || c > ‘9‘) { if(c == ‘-‘) k = -1; c = getchar();}
    while(c >= ‘0‘ && c <= ‘9‘) x = x * 10 + c - ‘0‘, c = getchar();
    return x * k;
}

bool cmp(node a, node b)
{
    if(a.t != b.t) return a.t < b.t;
    else return a.pl < b.pl;
}

bool cmp2(node a, node b)
{
    return a.pl < b.pl;
}

bool cmp3(node a, node b)
{
    return a.pl > b.pl;
}

void add(int x, int num)
{
    for(int i = x; i <= n; i += lowbit(i))
        c[i] += num;
}

ll query(int x)
{
    ll ans = 0;
    for(int i = x; i; i -= lowbit(i))
        ans += c[i];
    return ans;
}
  
void CDQ(int l, int r)//位置在我之前，num>我的 
{
    int mid = (l + r) >> 1;
    if(r - l >= 2) CDQ(l, mid), CDQ(mid + 1, r);
    if(l == r) return;
    sort(w + l, w + 1 + mid, cmp2);
    sort(w + mid + 1, w + r + 1, cmp2);
    int i = l, j = mid + 1;
    while(i <= mid && j <= r)
    {
        if(w[i].pl < w[j].pl) add(w[i].num, 1), i ++;
        else w[j].ans += (query(n) - query(w[j].num)), j ++; 
    }
    while(i <= mid) add(w[i].num, 1), i ++;
    while(j <= r) w[j].ans += (query(n) - query(w[j].num)), j ++;
    for(int i = l; i <= mid; i ++)
        add(w[i].num, -1);
}

void CDQ2(int l, int r)//位置在我之後，num<我的 
{
    int mid = (l + r) >> 1;
    if(r - l >= 2) CDQ2(l, mid), CDQ2(mid + 1, r);
    if(l == r) return;
    sort(w + l, w + 1 + mid, cmp3);
    sort(w + mid + 1, w + r + 1, cmp3);
    int i = l, j = mid + 1;
    while(i <= mid && j <= r)
    {
        if(w[i].pl > w[j].pl) add(w[i].num, 1), i ++;
        else w[j].ans += (query(w[j].num)), j ++; 
    }
    while(i <= mid) add(w[i].num, 1), i ++;
    while(j <= r) w[j].ans += (query(w[j].num)), j ++;
    for(int i = l; i <= mid; i ++)
        add(w[i].num, -1);
}


int main()
{
    n = read(), m = read();
    for(int i = 1; i <= n; i ++)
    {
        a[i] = read();
        b[a[i]] = i;
    }
    timer = m;
    for(int i = 1; i <= m; i ++)
    {
        d[i] = read();
        t[b[d[i]]] = timer --;
    }
    for(int i = 1; i <= n; i ++) w[i].t = t[i], w[i].num = a[i], w[i].pl = i;
    sort(w + 1, w + 1 + n, cmp);
    CDQ(1, n);
    for(int i = 1; i <= n; i ++) ans[w[i].t] += w[i].ans;
    for(int i = 1; i <= n; i ++) w[i].t = t[i], w[i].num = a[i], w[i].pl = i, w[i].ans = 0;
    sort(w + 1, w + 1 + n, cmp);
    memset(c, 0, sizeof(c));
    CDQ2(1, n);
    for(int i = 1; i <= n; i ++) ans[w[i].t] += w[i].ans;
    for(int i = 1; i <= m; i ++) ans[i] += ans[i - 1];
    for(int i = m; i >= 1; i --) printf("%lld\n", ans[i]);
    return 0;
}

【算法】CDQ分治 -- 三維偏序 & 動態逆序對