首页 > 代码库 > POJ2299 Ultra-QuickSort 【树状数组】+【hash】

POJ2299 Ultra-QuickSort 【树状数组】+【hash】

2024-07-11 21:51:26   219人阅读

Ultra-QuickSort
Time Limit: 7000MS Memory Limit: 65536K
Total Submissions: 39529 Accepted: 14250

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

注意结果要用long long存储。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#define maxn 500002
using std::sort;

int tree[maxn], ori[maxn], hash[maxn];
long long ans;

int getHash(int val, int n)
{
	int left = 0, right = n - 1, mid;
	while(left <= right){
		mid = (left + right) >> 1;
		if(val < hash[mid]) right = mid - 1;
		else if(val > hash[mid]) left = mid + 1;
		else return mid + 1;
	}
}

int lowBit(int pos){ return pos & (-pos); }

int getSum(int pos)
{
	int sum = 0;
	while(pos > 0){
		sum += tree[pos];
		pos -= lowBit(pos);
	}
	return sum;
}

void update(int pos, int n)
{
	ans += (pos - 1 - getSum(pos - 1));
	while(pos <= n){
		++tree[pos];
		pos += lowBit(pos);
	}
}


int main()
{
	int n, i;
	while(scanf("%d", &n), n){
		for(i = 0; i < n; ++i){
			scanf("%d", ori + i);
			hash[i] = ori[i];
		}
		
		sort(hash, hash + n);		
		for(i = 0; i < n; ++i)
			ori[i] = getHash(ori[i], n);			
		memset(tree, 0, sizeof(tree));
		
		ans = 0;
		for(i = 0; i < n; ++i) update(ori[i], n);
		printf("%lld\n", ans);
	}
	
	return 0;
}


联系
我们