Divide and Conquer(卜东波计算机算法作业题1)

Divide and Conquer
Recall the problem of finding the number of inversions. As in the course, we are given a sequence of n numbers a1, ... , a n, which we assume are all distinct, and we define an inversion to be a pair i< j such that a i>a j.
We motivated the problem of counting inversions as a good measure of how different two orderings are. However, one might feel that this measure is too sensitive. Let’s call a pair a significant inversion if i< j and a i> 3a j. Given an O(nlogn) algorithm to count the number of significant inversions between two orderings.
解：
3.1算法思想
采用分治法和归并排序思想，不停地将序列A分割为两个等长的子序列L 和R，分别对L和R中的significant inversions进行计数，然后对L和R组成的总序列A中所有的significant inversions进行合并计数。

A中的总计数为L和R 中的分别计数，加上由于L和R合并而引入的新计数。

当序列只有两个数的时候,就是递归的终止条件。

3.2算法
Sort-and-Count (A)
1: Divide A into two sub-sequence L and R;
2: (RC L, L) = Sort-and-Count (L);
3: (RC R, R) = Sort-and-Count (R);
4: (r, A) = Merge-and-Count (L, R);
5: return (RC = RC L + RC R + r, A);
Merge-and-Count (L, R)
1: InverseCount = 0;
2: i = 1; j = 1;
3: for k = 1 to r do
4: if L[i] > R[j] then
5: A[k] = L[j];
6: j + +;
7: else
8: A[k] = R[i];
9: i + +;
10: end if
11:end for
12:i = 1; j = 1;
13:for k = Ltor do
14: if L[i] > 3R[j] then
15: InverseCount+ = length(L) -i + 1;
16: + + j;
17: else
18: + + i;
19: end if
20:end for
21: return InverseCount and A;
3.3正确性证明：
For every i, we count the number of significant inversions betweena i and all a j’s. If a i<=3 a j, then there are no significant inversions between a j and any a m s.t. m>=j, so we decrease j. If a i>3a j, then a i>3a m for all m s.t. k<m<=j. In other words, we have detected j-k significant inversions involving a i.
3.4算法复杂度分析：
Merge-and-count中的每个for循环都最多执行n次，故Merge-and-count的时间复杂度为O（n），根据二分算法的特性，最多分割logn次，也就是Merge-and-count最多执行logn次，故本算法的总复杂度为O（nlogn）。