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poj 2299 Ultra-QuickSort(归并排序)


题目:​​http://poj.org/problem?id=2299​​


Ultra-QuickSort


Time Limit: 7000MS

 

Memory Limit: 65536K

Total Submissions: 49165

 

Accepted: 17984


Description




poj 2299 Ultra-QuickSort(归并排序)_i++

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


60

本来想用这题来练习线段树的,结果怎么也想不到那里去。后来看别人的代码才明白每个叶子节点存储的不是依据dex存储val[i]而是依据dex在相应位置存储1,然后根据排好序的序列由val[i]追踪dex,左边的1都加起来(逆序数),然后去掉dex的1,如此反复进行就能到达目的。提交代码时我可没这么做,用归并排序累加每一次相邻的交换数就是最终的结果。(临时在百度百科上学习了归并排序||-_-)


冒泡排序的时间的复杂度是O(n^2).

冒泡排序算法的运作如下:(从后往前)

比较相邻的元素。如果第一个比第二个大,就交换他们两个。

对每一对相邻元素作同样的工作,从开始第一对到结尾的最后一对。在这一点,最后的元素应该会是最大的数。

针对所有的元素重复以上的步骤,除了最后一个。

持续每次对越来越少的元素重复上面的步骤,直到没有任何一对数字需要比较。

并归排序的时间复杂度:O(nlogn)

设有数列{6,202,100,301,38,8,1}

初始状态:6,202,100,301,38,8,1

第一次归并后:{6,202},{100,301},{8,38},{1},比较次数:3;

第二次归并后:{6,100,202,301},{1,8,38},比较次数:4;

第三次归并后:{1,6,8,38,100,202,301},比较次数:4;

总的比较次数为:3+4+4=11,;

逆序数为14;

归并排序的优势:分治法,已比较了的小部分不再比较,如:{8,38},{1}变成{1,8,38}只比较1和8后{1}的长度成为0,不再比较,直接写下{8,38}。


#include <iostream>
#include <cstdio>
using namespace std;
const int maxn=5e5+10;
int a[maxn],b[maxn];
long long sum;
void Merge(int sourceArr[],int tempArr[],int startdex,int middex,int enddex)
{
int i = startdex,j=middex+1,k = startdex;
while(i<=middex && j<=enddex)
{
if(sourceArr[i]<sourceArr[j])
tempArr[k++] = sourceArr[i++];
else
tempArr[k++] = sourceArr[j++],sum+=(middex+1-i); //if sum++; not adjacent exchange
}
while(i!=middex+1)
tempArr[k++] = sourceArr[i++];
while(j!=enddex+1)
tempArr[k++] = sourceArr[j++];
for(i=startdex;i<=enddex;i++)
sourceArr[i] = tempArr[i];
}

void MergeSort(int sourceArr[],int tempArr[],int startdex,int enddex)
{
int middex;
if(startdex<enddex)
{
middex=(startdex+enddex)/2;
MergeSort(sourceArr,tempArr,startdex,middex);
MergeSort(sourceArr,tempArr,middex+1,enddex);
Merge(sourceArr,tempArr,startdex,middex,enddex);
}
}
int main()
{
//freopen("cin.txt","r",stdin);
int n;
while(cin>>n&&n){
sum=0;
for(int i=0;i<n;i++) scanf("%d",&a[i]);
MergeSort(a,b,0,n-1);
printf("%lld\n",sum);
}
return 0;
}




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