cdq分治
核心思想:在归并排序的过程中利用树状数组来统计答案(每次只分治右边对左边逆序对产生的贡献,其余递归处理即可,利用双指针的方法对于左边区间加入树状数组,对于右边区间query统计答案)
需要注意值相同时的处理方式
#include <iostream>
#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#define IOS ios_base:sync_with_stdio(0); cin.tie(0);
#define ll long long
#define INF 0x3f3f3f3f
#define MEM(x,y) memset(x,y,sizeof(x))
#define int long long
#define rep(i , a , b) for(int i = a ; i <= b ; i ++)
#define P pair<int,int>
#define sc(a) scanf("%lld",&a)
#define pf(a) printf("%lld ",a)
using namespace std;
const int N = 1000010, 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];
}
signed main()
{
cin >> n >> m;
for (int i = 0; i < n; i++)
{
int a, b, c;
cin >> 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]);
}

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