Problem B: Ultra-QuickSort

9 1 0 5 4 ,

0 1 4 5 9 .

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

思路：利用树状数组求逆序对。。但是数字很大的时候就需要离散化。

代码：

#include <stdio.h>
#include <string.h>
#include <map>
#include <algorithm>
using namespace std;

const int N = 500005;
int n, num[N], save[N], bit[N];
map<int, int> hash;

void add(int x, int v) {
	while (x <= n) {
		bit[x] += v;
		x += (x&(-x));
	}
}

int get(int x) {
	int ans = 0;
	while (x > 0) {
		ans += bit[x];
		x -= (x&(-x));
	}
	return ans;
}

void init() {
	hash.clear();
	memset(bit, 0, sizeof(bit));
	for (int i = 1; i <= n; i++) {
		scanf("%d", &num[i]);
		save[i] = num[i];
	}
	sort(save + 1, save + n + 1);
	for (int j = 1; j <= n; j++)
		hash[save[j]] = j;
}

long long solve() {
	long long ans = 0;
	for (int i = 1; i <= n; i++) {
		ans += get(n) - get(hash[num[i]]);
		add(hash[num[i]], 1);
	}
	return ans;
}

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

UVA 10810 - Ultra-QuickSort（树状数组+离散化)