10810 - Ultra-QuickSort（歸並排序求逆序數）

Problem B: Ultra-QuickSort

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of ndistinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence

9 1 0 5 4 ,

Ultra-QuickSort produces the output

0 1 4 5 9 .

Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Output for Sample Input

6
0

題意：大概就是求排好序需要交換幾次。。

思路：利用歸並排序在排序過程中求出逆序數。時間復雜度O(nlog)

代碼：

#include 
#include 

const int N = 500005;
int n, num[N], save[N], sn;

void init() {
	for (int i = 0; i < n; i++)
		scanf("%d", &num[i]);
}

long long solve(int l, int r, int *num) {
	if (r - l < 1) return 0;
	int mid = (l + r) / 2;
	long long ans = solve(l, mid, num) + solve(mid + 1, r, num);
	sn = l;
	int ll = l, rr = mid + 1;
	while (ll <= mid && rr <= r) {
		if (num[ll] <= num[rr])
			save[sn++] = num[ll++];
		else {
			ans += (mid + 1 - ll);
			save[sn++] = num[rr++];
		}
	}
	while (ll <= mid) save[sn++] = num[ll++];
	while (rr <= r) save[sn++] = num[rr++];
	for (int i = l; i <= r; i++)
		num[i] = save[i];
	return ans;
}

int main() {
	while (~scanf("%d", &n) && n) {
		init();
		printf("%lld\n", solve(0, n - 1, num));
	}
	return 0;
}