动态规划（二分法改进）：LIS问题以及应用题 （子序列型）

LIS 问题解决思路，着手dp数组记录下长度为b的单调递增子序列最小结束元素。

Example: array = [2,1,4,3,8,7,5,6,9]

对每个元素array[i]进行单独判断，考虑array[i]可构成最长为多长的局部LIS。

局部的搜索采用二分，以dp数组为域，以array[i]为判定值，找到一个dp中的位置 j 满足：

• ( dp[j]array[i] ) or dp[j]==NULL 

过程：O(n) in visit array, O(logn) in binsearch_dp -> O(nlogn)

• array = [2,1,4,3,8,7,5,6,9]
•     dp = [2,0,0,0,0,0,0,0,0]
•     dp = [1,0,0,0,0,0,0,0,0]
•     dp = [1,4,0,0,0,0,0,0,0]
•     dp = [1,3,0,0,0,0,0,0,0]
•     dp = [1,3,8,0,0,0,0,0,0]
•    ......
•     dp = [1,3,5,6,9,0,0,0,0]

可以得到最长长度是5,某一个LIS是 【1,3,5,6,9】

 C++实现代码：

int Longest_increasing_subseq(vector &array){
    vector dp(array.size());
    dp[0]=-1;
    int left,right,mid;
    int count=0;
    for(int i=0;idp[mid]){
                left=mid+1;
            }else{
                cout << "case Error happened!\n";
            }
        }
        if(dp[left]==0){
            dp[left]=array[i];
            ++count;
        }
        else{
            dp[left]=min(dp[left],array[i]);
        }
    }
    return count;
}

应用题：SlalomSkiing

You are a skier participating in a giant slalom. The slalom track is located on a ski slope, goes downhill and is fenced by barriers on both sides. The barriers are perpendicular to the starting line located at the top of the slope. There are N slalom gates on the track. Each gate is placed at a distinct distance from the starting line and from the barrier on the right-hand side (looking downhill).

You start from any place on the starting line, ski down the track passing as many gates as possible, and finish the slalom at the bottom of the slope. Passing a gate means skiing through the position of the gate.

You can ski downhill in either of two directions: to the left or to the right. When you ski to the left, you pass gates of increasing distances from the right barrier, and when you ski to the right, you pass gates of decreasing distances from the right barrier. You want to ski to the left at the beginning.

Unfortunately, changing direction (left to right or vice versa) is exhausting, so you have decided to change direction at most two times during your ride. Because of this, you have allowed yourself to miss some of the gates on the way down the slope. You would like to know the maximum number of gates that you can pass with at most two changes of direction.

The arrangement of the gates is given as an array A consisting of N integers, whose elements specify the positions of the gates: gate K (for 0 ≤ K < N) is at a distance of K+1 from the starting line, and at a distance of A[K] from the right barrier.

For example, consider array A such that:

A[0] = 15 A[1] = 13 A[2] = 5 A[3] = 7 A[4] = 4 A[5] = 10 A[6] = 12 A[7] = 8 A[8] = 2 A[9] = 11 A[10] = 6 A[11] = 9 A[12] = 3

 

 

The picture above illustrates the example track with N = 13 gates and a course that passes eight gates. After starting, you ski to the left (from your own perspective). You pass gates 2, 3, 5, 6 and then change direction to the right. After that you pass gates 7, 8 and then change direction to the left. Finally, you pass gates 10, 11 and finish the slalom. There is no possible way of passing more gates using at most two changes of direction.

Write a function:

class Solution { public int solution(int[] A); }

that, given an array A consisting of N integers, describing the positions of the gates on the track, returns the maximum number of gates that you can pass during one ski run.

For example, given the above data, the function should return 8, as explained above.

For the following array A consisting of N = 2 elements:

A[0] = 1 A[1] = 5

the function should return 2.

Write an efficient algorithm for the following assumptions:

• N is an integer within the range [1..100,000];
• each element of array A is an integer within the range [1..1,000,000,000];
• the elements of A are all distinct.

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 二重翻转数组，构造单调递增路径，用LIS解题：

int solution(vector &A){
    vector B;
    int mmax=(*max_element(A.begin(),A.end()))+1;
    // overturn A to two states, and extend to B as input_Array
    for (int i=0;i < A.size();++i){
        // avoid extend_elements destory A charactor of Increasing subseq in local
        B.push_back(mmax*2+A[i]);
        B.push_back(mmax*2-A[i]);
        B.push_back(A[i]);
    }
    return Longest_increasing_subseq(B);
}