Leetcode354. Two-dimensional nesting problem (converted to one-dimensional LIS)

Meaning:
given $The$ leader of group$n$ is$a_i$, The width is $b_i$The red envelope, if $a_i<a_j$且$b_i<b_j$, Then red envelope $i$ can be installed in red envelope$j$ . Now ask you how many red envelopes you can nest at most at a time.
Data range:$1\le n\le 10^5,1\le a_i,b_i\le 10^9$

Solution:
Consider the first dimension aa in ascending order$a$ and thenbbfor the second dimension$b$ Do LIS again. The problem here is that there may be the same$a$ red envelope nested within each other, which is not in line with the requirements, so we have a second dimension$b\; is$ sorted in descending order to ensure that each dimension$A$ can only be selected at most one.

Code:

const int N = 100010;
struct Node {
    
    
    int w, h;
    bool operator < (const Node &A) const {
    
    
        if(w == A.w) return h > A.h;
        return w < A.w;
    }
}a[N];
int t[N], tg;

int b_s(int l, int r, int x) {
    
    
    while(l < r) {
    
    
        int mid = l + r >> 1;
        if(a[t[mid]].h >= x) r = mid;
        else l = mid + 1;
    }
    return l;
}

class Solution {
    
    
public:
    int maxEnvelopes(vector<vector<int>>& env) {
    
    
        int n = 0;
        for(auto &u : env) a[++n] = {
    
    u[0], u[1]};
        if(n == 0) return 0;
        
        sort(a + 1, a + n + 1);
        
        tg = 0;
        t[++tg] = 1;
        for(int i = 2; i <= n; ++i) {
    
    
            if(a[i].h > a[t[tg]].h) t[++tg] = i;
            else if(a[i].h < a[t[tg]].h) t[b_s(1, tg, a[i].h)] = i;
        }
        
        return tg;
    }
};