Ultra-QuickSort （poj 2002)

 

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

这题很简单的样子，就是求冒泡排序的交换次数，but   超时

归并排序，求逆序数，别问我是什么？看着模板写就好

 1 #include<iostream>

 2 #include<cstring>

 3 #include<cstdio>

 4 using namespace std;

 5 int s[500005],temp[500005];

 6 long long cut;//这里害我WA了一次

 7 void merge_sort(int* A,int x,int y,int* T )

 8 {

 9     if(y-x>1)

10     {

11         int m=x+(y-x)/2;//划分

12         int p=x,q=m,i=x;

13         merge_sort(A,x,m,T);//递归求解

14         merge_sort(A,m,y,T);

15         while(p<m||q<y)

16         {

17             if(q>=y||(p<m&&A[p]<=A[q]))

18                 T[i++]=A[p++];//从左半数组复制到临时空间

19             else

20             {

21                 T[i++]=A[q++];//从右半数组复制到临时空间

22                 cut+= (m-p);//统计逆序数

23             }

24         }

25         for(i=x; i<y; i++)

26             A[i]=T[i];//从辅助数组复制回原数组

27     }

28 }

29 int main()

30 {

31     int n,i;

32     while(scanf("%d",&n)&&n)

33     {

34         cut=0;

35         for(i=0; i<n; i++)

36             scanf("%d",&s[i]);

37         merge_sort(s,0,n,temp);

38         printf("%lld\n",cut);

39     }

40     return 0;

41 }

View Code

 1 #include<iostream>

 2 using namespace std;

 3 long long  cnt;

 4 void merge(int array[],int left,int mid,int right)

 5 {

 6     int* temp=new int[right-left+1];

 7     int i,j,p;

 8     for(i=left,j=mid+1,p=0; i<=mid&&j<=right; p++)

 9     {

10         if(array[i]<=array[j])temp[p]=array[i++];

11         else temp[p]=array[j++],cnt+=(mid-i+1);

12     }

13     while(i<=mid)temp[p++]=array[i++];

14     while(j<=right)temp[p++]=array[j++];

15     for(i=left,p=0; i<=right; i++)array[i]=temp[p++];

16     delete temp;

17 }

18 void mergesort(int array[],int left,int right)

19 {

20     if(left==right)array[left]=array[right];

21     else

22     {

23         int mid=(left+right)/2;

24         mergesort(array,left,mid);

25         mergesort(array,mid+1,right);

26         merge(array,left,mid,right);

27     }

28 }

29 int main()

30 {

31     int n,array[500005];

32     while(cin>>n&&n)

33     {

34         cnt=0;

35         for(int i=0; i<n; i++)

36             cin>>array[i];

37         mergesort(array,0,n-1);

38         cout<<cnt<<endl;

39     }

40     return 0;

41 }

View Code

下面这个是网上找的还算好懂得，耗时是我敲得那个的10倍左右

Hint

Huge input and output,scanf and printf are recommended.