对于等边和等腰的一起处理，然后普通的三角形就利用双指针来处理

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int N = 5010;
struct node
{
    int len, sum;
    friend bool operator<(node x, node y)
    {
        return x.len < y.len;
    }
} a[N];

int main()
{
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        for (int i = 1; i <= n; ++i)
            cin >> a[i].len >> a[i].sum;
        sort(a + 1, a + 1 + n);
        ll ans = 0;
        for (int i = 1; i <= n; ++i)
            if (a[i].sum >= 2)
            {
                auto it = lower_bound(a + 1, a + 1 + n, (node){2 * a[i].len, 0});
                ll res = distance(a + 1, it); // 这个res是满足条件的最大解
                // 要区分res里面有没有i这个点，再判断能不能形成等边三角形
                if (i<=res && a[i].sum<3)
                --res;
                ans+=res;
            }
        for (int i = 1; i <= n - 2; ++i)
        {
            int k = i + 1;
            for (int j = i + 1; j <= n - 1; ++j)
            {
                while (k + 1 <= n && a[i].len + a[j].len > a[k + 1].len)
                    ++k;
                ans += k - j;
            }
        }
        cout << ans << "\n";
    }
}