【微软2014实习生及秋令营技术类职位在线测试】题目3 : Reduce inversion count

时间限制:10000ms

单点时限:1000ms

内存限制:256MB

Description

Find a pair in an integer array that swapping them would maximally decrease the inversion count of the array. If such a pair exists, return the new inversion count; otherwise returns the original inversion count.

Definition of Inversion: Let (A[0], A[1] ... A[n], n <= 50) be a sequence of n numbers. If i < j and A[i] > A[j], then the pair (i, j) is called inversion of A.

Example:

Count(Inversion({3, 1, 2})) = Count({3, 1}, {3, 2}) = 2

InversionCountOfSwap({3, 1, 2})=>

{

InversionCount({1, 3, 2}) = 1 <-- swapping 1 with 3, decreases inversion count by 1

InversionCount({2, 1, 3}) = 1 <-- swapping 2 with 3, decreases inversion count by 1

InversionCount({3, 2, 1}) = 3 <-- swapping 1 with 2 , increases inversion count by 1

}

Input

Input consists of multiple cases, one case per line.Each case consists of a sequence of integers separated by comma.

Output

For each case, print exactly one line with the new inversion count or the original inversion count if it cannot be reduced.

样例输入

3,1,2
1,2,3,4,5

样例输出

1
0

思路，此题采用的是暴力法，不过改进的一个地方是计算InversionCount的算法，采用的是合并排序，时间复杂度是O（nlogn）

import java.util.Scanner;

public class Main {

	static int InversionCount ;

	public static void main(String[] args)
	{
		int T,t;
		Scanner jin = new Scanner(System.in);
		while (jin.hasNext()) {
			String str = jin.next();
			String[] argstr = str.split(",");
			int[] num = new int[argstr.length];
			int[] tmp = new int[argstr.length];
			for (int i = 0; i < num.length; i++) {
				num[i] = Integer.parseInt(argstr[i]);
				tmp[i] = Integer.parseInt(argstr[i]);
			}
			InversionCount = 0;
			MergeSort(tmp, 0, tmp.length-1);
			int ret = InversionCount;
			for (int i = 0; i < num.length; i++)
				tmp[i] = num[i];
			for (int i = 0; i < num.length-1; i++) {
				for (int j = i+1; j < num.length; j++) {
					if (num[i] > num[j]) {
						int tmpnum = num[i];
						num[i] = num[j];
						num[j] = tmpnum;
						InversionCount = 0;
						MergeSort(num, 0, num.length-1);
						ret = Math.min(ret, InversionCount);
						for (int k = 0; k < num.length; k++)
							num[k] = tmp[k];
					}
				}
			}
			System.out.println(ret);
		}
	}

	public static void MergeSort(int[] array, int lhs, int rhs) {
		if (lhs < rhs) {
			int mid = lhs + ((rhs - lhs)>>1);
			MergeSort(array, lhs, mid);
			MergeSort(array, mid+1, rhs);
			Merge(array, lhs, mid, rhs);
		}
	}
	public static void Merge(int[] array, int lhs, int mid, int rhs) {
		int[] tmp = new int[rhs-lhs+1];
		int i = lhs, j = mid+1;
		int k = 0;
		while(i <= mid && j <= rhs)
		{
			if (array[i] > array[j]) {
				InversionCount += mid-i+1;
				tmp[k++] = array[j++];
			}
			else {
				tmp[k++] = array[i++];
			}
		}
		while(i <= mid)
		{
			tmp[k++] = array[i++];
		}
		while(j <= rhs)
		{
			tmp[k++] = array[j++];
		}
		for (i = 0; i < k; i++) {
			array[i+lhs] = tmp[i];
		}
		tmp = null;
	}

}

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