GYM 101889F（树状数组）

bit扫描坐标套路题，注意有重复的点，莽WA了。

const int maxn = 1e5 + 5;
struct node {
    ll B, F, D;

    bool operator < (const node &rhs) const {
        if (B != rhs.B) return B > rhs.B;
        return F < rhs.F;
    }
}a[maxn];
int n, tot;
ll c[maxn], ans;
struct BIT {
    ll t[maxn];

    void Modify(int x, ll val) {
        for (; x; x -= x&-x)
            t[x] = max(t[x], val);
    }

    ll Query(int x) {
        ll ret = 0;
        for (; x <= tot; x += x&-x)
            ret = max(t[x], ret);
        return ret;
    }
}T;

int main() {
    read(n);
    rep(i, 1, n) {
        read(a[i].B);
        read(a[i].F);
        read(a[i].D);
        c[++tot] = a[i].F;
    }

    sort(c + 1, c + tot + 1);
    tot = unique(c + 1, c + 1 + tot) - c - 1;
    rep(i, 1, n) {
        a[i].F = lower_bound(c + 1, c + 1 + tot, a[i].F) - c;
    }
    sort(a + 1, a + 1 + n);

    ll tmp = 0;
    rep(i, 1, n) {
        if (a[i].B == a[i - 1].B && a[i].F == a[i - 1].F)
            tmp += a[i].D;
        else    tmp = T.Query(a[i].F + 1) + a[i].D;
        ans = max(ans, tmp);
        T.Modify(a[i].F, tmp);
    }
    writeln(ans);
    return 0;
}