【 樹狀數組 】POJ 2299 Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 66429		Accepted: 24883

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct 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.

Input

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.

Output

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

Sample Output

6
0

求逆序數，經典

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 500005;
int n;
int tree[maxn];
int a[maxn];
struct node{
	int val,pos;
}p[maxn];

bool cmp(node a, node b){
	return a.val < b.val;
}

int lowbit(int x) {	return x&(-x); }

void Add(int a, int b)
{
	while(a <= n){
		tree[a] += b;
		a += lowbit(a);
	}
}

int Query(int x)
{
	int ans = 0;
	while(x>0){
		ans += tree[x];
		x -= lowbit(x);
	}
	return ans;
}

int main()
{
	while(~scanf("%d",&n))
	{
		if(n==0) break;
		memset(tree,0,sizeof tree);
		for(int i = 1,x; i <= n; i++){
			scanf("%d",&p[i].val);
			p[i].pos = i;
		}
		sort(p+1,p+n+1,cmp);
		for(int i = 1; i <= n; i++){ //離散化
			a[p[i].pos] = i;
			//printf("%d   ",p[i].pos);
		}
		printf("\n");
		long long ans = 0;
		for(int i = 1; i <= n; i++){
			Add(a[i],1);
			ans += (i - Query(a[i]));
		}
		printf("%lld\n",ans);
	}
}