leetcode 354. Russian Doll Envelopes

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

其实这道题目就是最长上升子序列(LIS)的改编。
思路：先固定一维，然后在另一维进行求LIS操作。注意在固定一维的时候，可能存在(12,6)(12,5)(1,3)怎么排序的问题，对于这种情况，要把第二维大的排在前面如(1,3)(12,6)(12,5)。因为只有这样，对于第二维求LIS才能保证正确性(就是贪心嘛，第一维相同那么我就取第二维比较大的那个，因为只有这样才能让小的有机会替换，回想LIS中的d序列...)。

class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
        int n = envelopes.size();
        if (n == 0) return 0;
        if (n == 1) return 1;
        auto cmp = [](pair<int, int> a, pair<int, int> b)   
        { return a.first == b.first ? a.second > b.second : a.first < b.first; };  
        sort(envelopes.begin(), envelopes.end(), cmp);

        vector<int> p(n+1);
        p[0] = envelopes[0].second;
        int len = 0;
        
        for (int i = 1; i < n; ++i) {
            if (envelopes[i].second > p[len]) {
                ++len;
                p[len] = envelopes[i].second;
            } else if (envelopes[i].second < p[len]) {
                int pos = lower_bound(p.begin(), p.begin()+len, envelopes[i].second) - p.begin();
                p[pos] = envelopes[i].second;
            }
        }
        return len + 1;
    }
};