一、本文内容
最长递增子序列的两种动态规划算法实现，O(n^2)及O(nlogn).
二、问题描述
最长递增子序列：给定一个序列，从该序列找出最长的 升序/递增 子序列。
特点：1、子序列不要求连续； 2、子序列在原序列中按严格(strictly)升序排序； 3、最长递增子序列不唯一。
注：下文最长递增子序列用缩写LIS表示。
example：
0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
对应的LIS：
0, 2, 6, 9, 13, 15
0, 4, 6, 9, 11, 15
三、算法描述
1、考察第i+1个元素时，不考虑前面i个元素的状态
给定长度为n的序列A[0..n-1]，对于给定某个的阶段状态，它以前各阶段的状态无法直接影响它未来的决策，而只能间接通过当前状态来影响；换句话说，每个状态都是过去历史的完整总结。即LIS[i]当前的状态与LIS[0..i-1]已无关。这几句话可以用Fibonacci的递归树来验证，对LIS的递归树同样适用。 如下面的递归树：
LIS(3)
/ | \
LIS(2) LIS(1) LIS(0)
/ \ /
LIS(1) LIS(0) LIS(0)
/
LIS(0)
很显然，递归树有很多重复/重叠的子问题，而大问题的最优解又可以由这些子问题的最优解得到，符合DP的两个条件，故可以用DP来解决LIS问题。
那么给定序列A[0..n-1]，求它的LIS，可以top-down分解任务，得到递归树；递归树bottom-up就可以建立子问题的查询表以供求解当前问题时查询。
这就是DP的两种处理方式：Memoization（top-down) 和 Tabulation (bottom-up).
假设LIS[i]表示A[0..i]的最长递增子序列的长度，那么
LIS[i+1] = max{1, LIS[j]+1}, A[i+1]>A[j], for any j<=i； 注意：到A[i]的子序列长度不一定大于A[j]的子序列长度（如上例当A[j]=12, LIS[j]={0, 4, 12}和当A[i]=2, LIS[i]={0, 2}）
若A[i+1]>A[j],则A[i+1]可以接在LIS[j]长的子序列后面构成一个更长的子序列。
同时, 从A[i+1]开始又构成一个长度为1的子序列。故两者取大值。
2、考察第i+1个元素时，考虑前面i个元素的状态
那么，什么时候在一个已存在的序列中添加或者替换一个元素是安全的呢？（DP算法都是offline algorithm，需要全局考虑）
显然，我们需要维护所有递增的子序列（称其为active list，即可能成为max{LIS}的子序列）。而这些子序列的长度都不同，
可以按照插入排序的思想，一个一个元素地从前面找到其应该归属的子序列(active list)。
有A[i]>A[i-1]和A[i]<A[i-1]两种情况，A[i]<A[i-1]又有一种特殊情况A[i]是当前序列中的最小元素，即A[i]<A[j], for any j<i;
case 1. If A[i] is smallest among all end candidates of active lists, we will start new active list of length 1.
（when we encounter new smallest element in the array, it can be a potential candidate to start new sequence，{2, 5, 3, 1, 2, 3, 4, 5, 6}）
case 2. If A[i] is largest among all end candidates of active lists, we will clone the largest active list, and extend it by A[i].
case 3. If A[i] is in between, we will find a list with largest end element that is smaller than A[i]. Clone and extend this list by A[i]. We will discard all other lists of same length as that of this modified list.
（找到end element小于A[i]的active list之后，其他相同长度的active list将被删除，因为A[i]小于这些等长active list的end element，用新的active list代替将被删除的active list：复制小于A[i]的active list，并添加A[i]）
处理过程如下
example：A[ ] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}
A[0] = 0. Case 1. There are no active lists, create one.
0.
-----------------------------------------------------------------------------
A[1] = 8. Case 2. Clone and extend.
0.
0, 8.
-----------------------------------------------------------------------------
A[2] = 4. Case 3. Clone, extend and discard.
0.
0, 4.
0, 8. Discarded
-----------------------------------------------------------------------------
A[3] = 12. Case 2. Clone and extend.
0.
0, 4.
0, 4, 12.
-----------------------------------------------------------------------------
A[4] = 2. Case 3. Clone, extend and discard.
0.
0, 2.
0, 4. Discarded.
0, 4, 12.
-----------------------------------------------------------------------------
A[5] = 10. Case 3. Clone, extend and discard.
0.
0, 2.
0, 2, 10.
0, 4, 12. Discarded.
-----------------------------------------------------------------------------
A[6] = 6. Case 3. Clone, extend and discard.
0.
0, 2.
0, 2, 6.
0, 2, 10. Discarded.
-----------------------------------------------------------------------------
A[7] = 14. Case 2. Clone and extend.
0.
0, 2.
0, 2, 6.
0, 2, 6, 14.
-----------------------------------------------------------------------------
A[8] = 1. Case 3. Clone, extend and discard.
0.
0, 1.
0, 2. Discarded.
0, 2, 6.
0, 2, 6, 14.
-----------------------------------------------------------------------------
A[9] = 9. Case 3. Clone, extend and discard.
0.
0, 1.
0, 2, 6.
0, 2, 6, 9.
0, 2, 6, 14. Discarded.
-----------------------------------------------------------------------------
A[10] = 5. Case 3. Clone, extend and discard.
0.
0, 1.
0, 1, 5.
0, 2, 6. Discarded.
0, 2, 6, 9.
-----------------------------------------------------------------------------
A[11] = 13. Case 2. Clone and extend.
0.
0, 1.
0, 1, 5.
0, 2, 6, 9.
0, 2, 6, 9, 13.
-----------------------------------------------------------------------------
A[12] = 3. Case 3. Clone, extend and discard.
0.
0, 1.
0, 1, 3.
0, 1, 5. Discarded.
0, 2, 6, 9.
0, 2, 6, 9, 13.
-----------------------------------------------------------------------------
A[13] = 11. Case 3. Clone, extend and discard.
0.
0, 1.
0, 1, 3.
0, 2, 6, 9.
0, 2, 6, 9, 11.
0, 2, 6, 9, 13. Discarded.
-----------------------------------------------------------------------------
A[14] = 7. Case 3. Clone, extend and discard.
0.
0, 1.
0, 1, 3.
0, 1, 3, 7.
0, 2, 6, 9. Discarded.
0, 2, 6, 9, 11.
----------------------------------------------------------------------------
A[15] = 15. Case 2. Clone and extend.
0.
0, 1.
0, 1, 3.
0, 1, 3, 7.
0, 2, 6, 9, 11.
0, 2, 6, 9, 11, 15. <-- LIS List
----------------------------------------------------------------------------
注：观察上面的处理过程，我们都只在处理所有active list的最后一个元素end element(粗体)，那么仅仅需要维护所有active list构成的end element集合（粗体斜边），
可以用一维数组来存储。discard操作可以用replace操作来模拟。
四、算法实现
第三节的算法描述序号分别对应下面的算法实现
1、时间复杂度为O(n^2)的DP实现
/**
@description: Longest Increasing Subsequence
@author: seiyagoo
@create: 2013.10.25
@modified: 2013.10.26
**/
int LIS_1(int A[], int size){ int *LIS = new int[size];
vector<int> *vec = new vector<int>[size]; /* Compute optimized LIS values in bottom up manner */
for(int i=; i < size; i++){
LIS[i]=; //初始化默认长度
int max_j=, flag=;
for(int j=; j < i; j++){ //查表,找出前面最长的序列, 若将A[i]加入LIS[j](LIS[j]+1的含义)的递增子序列比当前的LIS[i]更长, 则更新LIS[i]
if(A[i] > A[j] && LIS[i] < LIS[j]+){
LIS[i] = LIS[j]+;
max_j=j;
flag=;
}
}
if(flag) //copy前面最长子序列到vec[i]
vec[i].insert(vec[i].end(), vec[max_j].begin(), vec[max_j].end());
vec[i].push_back(A[i]); //最后放入A[i]
} /*Show LIS of the current state*/
vector<int>::iterator it;
cout<<left;
for(int i=; i<size; i++){
cout<<setw()<<A[i]<< " --> ";
for(it = vec[i].begin(); it!=vec[i].end(); it++)
cout<<*it<<" ";
cout<<endl;
} /* Pick maximum of all LIS values, namely max{LIS[i]} */
int max_len=;
for(int i = ; i < size; i++ )
if( max_len < LIS[i] )
max_len = LIS[i]; delete[] LIS;
delete[] vec; return max_len;
}
2、时间复杂度为O(nlogn)的DP实现
/**
@description: Longest Increasing Subsequence
@author: seiyagoo
@create: 2013.10.25
@modified: 2013.10.26
**/ // Binary search (note boundaries in the caller)
// A[] is ceilIndex in the caller
int CeilIndex(int A[], int l, int r, int key) {
int m; while( r - l > ) {
m = l + (r - l)/;
(A[m] >= key ? r : l) = m; // ternary expression returns an l-value
} return r;
} int LIS_2(int A[], int size) {
// boundary case: when array size is one
if( == size ) return ; int *tailTable = new int[size];
vector<int> *vec = new vector<int>[size];
int len; // always points empty slot //memset(tailTable, INT_MAX, sizeof(tailTable[0])*size); @bug for(int i = ; i < size; i++)
tailTable[i] = INT_MAX; tailTable[] = A[]; //tailTable[0] store the smallest value
vec[].push_back(A[]); len = ;
for( int i = ; i < size; i++ ) {
if( A[i] < tailTable[] ) { //case 1: new smallest value
tailTable[] = A[i]; /*discard and create*/
vec[].clear();
vec[].push_back(A[i]);
}
else if( A[i] > tailTable[len-] ) { //case 2: A[i] wants to extend largest subsequence
tailTable[len++] = A[i]; /*clone and extend*/
vec[len-] = vec[len-];
vec[len-].push_back(A[i]);
}
else { //case 3: A[i] wants to be current end candidate of an existing subsequence, It will replace ceil value in tailTable
int ceilIndex = CeilIndex(tailTable, -, len-, A[i]);
tailTable[ceilIndex] = A[i]; /*discard, clone and extend*/
vec[ceilIndex].clear();
vec[ceilIndex] = vec[ceilIndex-];
vec[ceilIndex].push_back(A[i]);
} /*Printf all the active lists*/
vector<int>::iterator it;
cout<<left;
cout<<"A["<<i<<"] = "<<A[i]<<endl<<endl;
cout<<"active lists:"<<endl;
for(int i=; i<len; i++){
for(it = vec[i].begin(); it!=vec[i].end(); it++)
cout<<*it<<" ";
cout<<endl;
} /*Printf end elements of all the active lists*/
cout<<endl<<"end elements array:"<<endl;
for(int i = ; i < size; i++)
if(tailTable[i] != INT_MAX)
cout<<tailTable[i]<<" ";
cout<<endl;
cout<<"-------------------------"<<endl;
} delete[] tailTable;
delete[] vec; return len;
}
五、运行结果
example：
0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15
算法实现一
算法实现二
参考：
《编程之美》