扫描线思路，每遇到一个区间左端点，这个区间就与之前出现的且还没结束的所有区间相交。

样例全过，WA #12，求调玄关

#include<bits/stdc++.h>
#define int long long 
using namespace std;

const int N = 5e5 + 5;
int n, sum, D[N * 2], cnt;
bool vis[N * 2];
long long ans;

struct node
{
	int l, r;
}a[N];

struct haha
{
	int l, r;
	vector<int> whl, whr;
}ha[N * 2];

signed main()
{
	cin >> n;
	for(int i = 1; i <= n; i++)
	{
		cin >> a[i].l >> a[i].r;
		D[i] = a[i].l;
		D[n + i] = a[i].r;
	}
	sort(D + 1, D + n * 2);
	cnt = unique(D + 1, D + 1 + n * 2) - (D + 1);
	for(int i = 1; i <= n; i++)
	{
		a[i].l = lower_bound(D + 1, D + 1 + cnt, a[i].l) - D;
		a[i].r = lower_bound(D + 1, D + 1 + cnt, a[i].r) - D;
		ha[a[i].l].l++;
		ha[a[i].r].r++;
	}
	for(int i = 1; i <= cnt; i++)
	{
		if(ha[i].l)
		{
			for(int j = 1; j <= ha[i].l; j++)
			{
				ans += sum;
				sum++;
			}
		}
		if(ha[i].r)
		{
			for(int j = 1; j <= ha[i].r; j++)
				sum--;
		}
	}
	cout << ans;
	return 0;
}