11.7 A circus is designing a tower routine consisting of people standing atop one another's shoulders. For practical and aesthetic reasons, each person must be both shorter and lighter than the person below him or her. Given the heights and weights of each person in the circus, write a method to compute the largest possible number of people in such a tower.
 EXAMPLE:
 Input (ht,wt): (65, 100) (70, 150) (56, 90) (75, 190) (60, 95) (68, 110)
 Output:The longest tower is length 6 and includes from top to bottom: (56, 90) (60,95) (65,100) (68,110) (70,150) (75,190)

這道題說馬戲團有一種人塔，上面的人要比下面的人既矮又輕，問我們最多能有多少個人組成人塔。那麼就相當於求最長的遞增子序列，我們的做法是先將所有的人按身高排個序，方法可參見我之前的部落格C++ sort vector<vector<int> > or vector<MyClass> 容器的排序。然後對於體重序列，就相當於找最長連續遞增子序列Longest Increasing Subsequence (LIS)，找LIS的方法可參見我之前的部落格Longest Increasing Subsequence 最長遞增子序列，參見程式碼如下：

class
 
 HtWt {
public:
    int Ht, Wt;
    HtWt(int h, int w): Ht(h), Wt(w) {}
    bool isBefore(HtWt *other) {
        if (Ht < other->Ht && Wt < other->Wt) return true;
        else return false;
    }
};

bool cmp(HtWt const *a, HtWt const *b) {
    return a->Ht < b->Ht;
}

 
class Solution {
public:
    vector<HtWt*> getIncreasingSequence(vector<HtWt*> &items) {
        sort(items.begin(), items.end(), cmp);
        return longestIncreasingSubsequence(items);
    }
    vector<HtWt*> longestIncreasingSubsequence(vector<HtWt*> &array) {
        vector<vector<HtWt*> > solutions;
        longestIncreasingSubsequence(array, solutions, 0);
        vector<HtWt*> res;
        for (auto &a : solutions) {
            res = seqWithMaxLength(res, a);
        }
        return res;
    }
    void longestIncreasingSubsequence(vector<HtWt*> &array, vector<vector<HtWt*> > &solutions, int curIdx) {
        if (curIdx >= array.size() || curIdx < 0) return;
        HtWt *cur = array[curIdx];
        vector<HtWt*> res;
        for (int i = 0; i < curIdx; ++i) {
            if (array[i]->isBefore(cur)) {
                res = seqWithMaxLength(res, solutions[i]);
            }
        }
        vector<HtWt*> new_solution = res;
        new_solution.push_back(cur);
        solutions.push_back(new_solution);
        longestIncreasingSubsequence(array, solutions, curIdx + 1);
    }
    vector<HtWt*> seqWithMaxLength(vector<HtWt*> seq1, vector<HtWt*> seq2) {
        if (seq1.empty()) return seq2;
        if (seq2.empty()) return seq1;
        return seq1.size() > seq2.size() ? seq1 : seq2;
    }
};