Related: Circus 找最大序列
[CareerCup] 11.7 Tower of People in Circus 马戏团的人塔 - Grandyang - 博客园
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)
这道题说马戏团有一种人塔，上面的人要比下面的人既矮又轻，问我们最多能有多少个人组成人塔。那么就相当于求最长的递增子序列，我们的做法是先将所有的人按身高排个序，方法可参见我之前的博客C++ sort vector<vector<int> > or vector<MyClass> 容器的排序。然后对于体重序列，就相当于找最长连续递增子序列Longest Increasing Subsequence (LIS)，找LIS的方法可参见我之前的博客Longest Increasing Subsequence 最长递增子序列
最长上升子序列，对高度排序然后对排序后的数组X找出按体重的最长上升子序列即可。时间复杂度O(nlogn)。
楼主这个解法也是正确的，可以证明AB的任一公共子序列都是X的上升子序列，且X的任一上升子序列都是AB的公共子序列。但是这个解法需要O(n^2)。
http://www.cnblogs.com/grandyang/p/4891506.html
[CareerCup] 11.7 Tower of People in Circus 马戏团的人塔 - Grandyang - 博客园
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)
这道题说马戏团有一种人塔，上面的人要比下面的人既矮又轻，问我们最多能有多少个人组成人塔。那么就相当于求最长的递增子序列，我们的做法是先将所有的人按身高排个序，方法可参见我之前的博客C++ sort vector<vector<int> > or vector<MyClass> 容器的排序。然后对于体重序列，就相当于找最长连续递增子序列Longest Increasing Subsequence (LIS)，找LIS的方法可参见我之前的博客Longest Increasing Subsequence 最长递增子序列
有两种方法：
1. 按照高度排序，为A
2. 按照体重排序，为B
3. 求A和B的最长公共子序列
第二种方法：
按照高度排序，然后找体重的最长上升序列
https://www.cnblogs.com/grandyang/p/4891506.html最长上升子序列，对高度排序然后对排序后的数组X找出按体重的最长上升子序列即可。时间复杂度O(nlogn)。
楼主这个解法也是正确的，可以证明AB的任一公共子序列都是X的上升子序列，且X的任一上升子序列都是AB的公共子序列。但是这个解法需要O(n^2)。
cut out all the "fluff" to this problem
We have a list of pairs of items. Find the longest sequence such that both the first and second items are in nondecreasing
http://www.cnblogs.com/chkkch/archive/2012/10/29/2744557.html6 struct Node 7 { 8 int height; 9 int weight; 10 11 Node(){} 12 Node(int h, int w):height(h), weight(w){} 13 }; 14 15 bool comp(const Node &lhs, const Node &rhs) 16 { 17 return lhs.height < rhs.height; 18 } 19 20 int solve(vector<Node> &a) 21 { 22 sort(a.begin(), a.end(), comp); 23 24 vector<int> f(a.size()); 25 26 f[0] = 1; 27 28 for(int i = 1; i < a.size(); i++) 29 { 30 f[i] = 1; 31 for(int j = 0; j < i; j++) 32 if (a[i].height > a[j].height && a[i].weight > a[j].weight) 33 f[i] = max(f[i], f[j] + 1); 34 } 35 36 int maxF = 0; 37 for(int i = 0; i < f.size(); i++) 38 maxF = max(maxF, f[i]); 39 40 return maxF; 41 }http://jianliu.org/2015/01/cc150-q11-7/
public class HtWt implements Comparable {
public int Ht;
public int Wt;
HtWt(int Ht, int Wt){
this.Ht = Ht;
this.Wt = Wt;
}
//override the compareTo method
public int compareTo(Object s){
HtWt second = (HtWt)s;
if(this.Ht!=second.Ht){
return ((Integer)this.Ht).compareTo((Integer)second.Ht);
}else{
return ((Integer)this.Wt).compareTo((Integer)second.Wt);
}
}
public boolean isBefore(HtWt other){
if(this.Ht <= other.Ht&& this.Wt<= other.Wt ){
return true;
}else{
return false;
}
}
public static ArrayList<HtWt> getSequence(ArrayList<HtWt> items){
//sort all the people according to their height
Collections.sort(items);
return findLSA(items);
}
//sort to find the longest sub array according to weight
public static ArrayList<HtWt> findLSA(ArrayList<HtWt> items){
//new the solution container
ArrayList<HtWt>[] solutions = new ArrayList[items.size()];
//fill in the solution array
for(int i=0;i<items.size();i++){
ArrayList<HtWt> temp = null;
//iterate all the solution array before i, find the arraylist which is the longest that can be appended
for(int j=0;j<i;j++){
//make sure it can append
if(items.get(j).isBefore(items.get(i))){
temp = MaxLength(temp,solutions[j]);
}
}
//deep copy the temp
ArrayList<HtWt> newTemp = new ArrayList<HtWt>();
if(temp!=null){
newTemp.addAll(temp);
}
newTemp.add(items.get(i));
solutions[i] = newTemp;
}
ArrayList<HtWt> result = solutions[0];
for(int i=1;i<solutions.length;i++){
result = MaxLength(solutions[i],result);
}
return result;
}
//helper method
public static ArrayList<HtWt> MaxLength(ArrayList<HtWt> seq1, ArrayList<HtWt> seq2){
if(seq1 == null){
return seq2;
}
if(seq2 == null){
return seq1;
}
return seq1.size()>seq2.size()?seq1:seq2;
}
http://blog.csdn.net/navyifanr/article/details/20837459http://www.cnblogs.com/grandyang/p/4891506.html
X. 2^n, brute force
public static ArrayList<HtWt> longestIncreasingSeq(ArrayList<HtWt> items) {
Collections.sort(items);
return bestSeqAtIndex(items, new ArrayList<HtWt>(), 0);
}
// Returns longer sequence
private static ArrayList<HtWt> max(ArrayList<HtWt> seq1, ArrayList<HtWt> seq2) {
if (seq1 == null) {
return seq2;
} else if (seq2 == null) {
return seq1;
}
return seq1.size() > seq2.size() ? seq1 : seq2;
}
private static boolean canAppend(ArrayList<HtWt> solution, HtWt value) {
if (solution == null) {
return false;
}
if (solution.size() == 0) {
return true;
}
HtWt last = solution.get(solution.size() - 1);
return last.isBefore(value);
}
private static ArrayList<HtWt> bestSeqAtIndex(ArrayList<HtWt> array, ArrayList<HtWt> sequence, int index) {
if (index >= array.size()) return sequence;
HtWt value = array.get(index);
ArrayList<HtWt> bestWith = null;
if (canAppend(sequence, value)) {
ArrayList<HtWt> sequenceWith = (ArrayList<HtWt>) sequence.clone();
sequenceWith.add(value);
bestWith = bestSeqAtIndex(array, sequenceWith, index + 1);
}
ArrayList<HtWt> bestWithout = bestSeqAtIndex(array, sequence, index + 1);
return max(bestWith, bestWithout);
}
