poj-----Ultra-QuickSort(离散化+树状数组)

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 38258		Accepted: 13784

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

Source

Waterloo local 2005.02.05

 

代码：

 1 //离散化+树状数组

 2 #include<stdio.h>

 3 #include<stdlib.h>

 4 #include<string.h>

 5 #define maxn 500000

 6 struct node

 7 {

 8     int val;

 9     int id;

10 };

11 node stu[maxn+5];

12 int n;

13 __int64 cnt;

14 int bb[maxn+5];

15 int lowbit(int x)

16 {

17   return x&(-x);

18 }

19 void ope(int x)

20 {

21     while(x<=n)

22     {

23      bb[x]++;

24      x+=lowbit(x);

25     }

26 }

27 int sum(int x)

28 {

29     int ans=0;

30     while(x>0)

31     {

32      ans+=bb[x];

33      x-=lowbit(x);

34     }

35     return ans;

36 }

37 int cmp(const void *a ,const void *b)

38 {

39   return (*(node *)a).val -(*(node *)b).val;

40 }

41 int main()

42 {

43     int i;

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

45     {

46         memset(bb,0,sizeof(int)*(n+1));

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

48         {

49          scanf("%d",&stu[i].val);

50          stu[i].id=i+1;

51         }

52         qsort(stu,n,sizeof(stu[0]),cmp);

53         cnt=0;

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

55         {

56             cnt+=sum(n)-sum(stu[i].id);

57             ope(stu[i].id);

58         }

59         printf("%I64d\n",cnt);

60     }

61   return 0;

62 }

 归并排序：

代码：

 1 //归并排序求逆序数

 2 #include<stdio.h>

 3 #include<string.h>

 4 #include<stdlib.h>

 5 #define maxn 500000

 6 int cc[maxn+5];

 7 int aa[maxn+5];

 8 __int64 cnt;

 9 void merge(int low ,int mid,int hig)

10 {

11     int i,j,k;

12     i=low;

13     j=mid;

14     k=0;

15     while(i<mid&&j<hig)

16     {

17         if(aa[i]>aa[j])

18         {

19             cc[k++]=aa[j++];

20             cnt+=mid-i;

21         }

22         else

23             cc[k++]=aa[i++];

24     }

25     while(i<mid)

26         cc[k++]=aa[i++];

27     while(j<hig)

28         cc[k++]=aa[j++];

29     for(k=0,i=low;i<hig;i++)

30         aa[i]=cc[k++];

31 }

32 void merge_sort(int st,int en)

33 {

34     int mid;

35     if(st+1<en)

36     {

37         mid=st+(en-st)/2;

38         merge_sort(st,mid);

39         merge_sort(mid,en);

40         merge(st,mid,en);

41     }

42 }

43 int main()

44 {

45     int n,i;

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

47     {

48         cnt=0;

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

50          scanf("%d",aa+i);

51          merge_sort(0,n);

52          printf("%I64d\n",cnt);

53     }

54     return 0;

55 }

View Code

 非递归归并排序：
代码：

 1 //归并排序求逆序数

 2 #include<stdio.h>

 3 #include<string.h>

 4 #include<stdlib.h>

 5 #define maxn 500000

 6 int cc[maxn+5];

 7 int aa[maxn+5];

 8 __int64 cnt;

 9 void merge(int low ,int mid,int hig)

10 {

11     int i,j,k;

12     i=low;

13     j=mid;

14     k=0;

15     while(i<mid&&j<hig)

16     {

17         if(aa[i]>aa[j])

18         {

19             cc[k++]=aa[j++];

20             cnt+=mid-i;

21         }

22         else

23             cc[k++]=aa[i++];

24     }

25     while(i<mid)

26         cc[k++]=aa[i++];

27     while(j<hig)

28         cc[k++]=aa[j++];

29     for(k=0,i=low;i<hig;i++)

30         aa[i]=cc[k++];

31 }

32 //void merge_sort(int st,int en)

33 //{

34 //    int mid;

35 //    if(st+1<en)

36 //    {

37 //        mid=st+(en-st)/2;

38 //        merge_sort(st,mid);

39 //        merge_sort(mid,en);

40 //        merge(st,mid,en);

41 //    }

42 //}

43 void merge_sort(int st,int en)

44 {

45     int s,t,i;

46     t=1;

47     while(t<=(en-st))

48     {

49         s=t;

50         t<<=1;

51         i=st;

52         while(i+t<=en)

53         {

54             merge(i,i+s,i+t);

55             i+=t;

56         }

57         if(i+s<en)

58             merge(i,i+s,en);

59     }

60       if(i+s<en)

61             merge(i,i+s,en);

62 }

63 int main()

64 {

65     int n,i;

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

67     {

68         cnt=0;

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

70          scanf("%d",aa+i);

71          merge_sort(0,n);

72          printf("%I64d\n",cnt);

73     }

74     return 0;

75 }

View Code