HUDOJ-----1394Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9163    Accepted Submission(s): 5642

Problem Description

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

 

 

Input

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

 

 

Output

For each case, output the minimum inversion number on a single line.

 

 

Sample Input

10 1 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16

 

 

Author

CHEN, Gaoli

 

 

Source

ZOJ Monthly, January 2003

 

 

       求逆序数，这道题花了我一下午的时间去看线代，不过还好总算做出了....切克闹，切脑壳...

下 面来详细讲讲过程吧...

        首先，我求出了 Simple output 给出的 序列的 逆序数为22 这是没有错的，但是输出却为16，当时我这个小脑袋呀，真是....泪崩了呀！. 

 然后我就在这里纠结呀...哎，由于英语不是很好，居然没有读懂这句话的意思.....这是啥情况 ，妈蛋呀！

     out of the above sequences.  ------>从上面的式子中找出最小的逆序数...

     明白了这句话，下面就好办了..

    最后就是一点要说的是... 对于逆序数，如果存在左右顶端对调，并且这个序列是连续的..

是可以总结出规律的,,

看代码就知道了。。

                             time   300+ms....   c++ 

 1 #include<stdio.h>

 2 #include<stdlib.h>

 3 #include<string.h>

 4 #define maxn 5000

 5 int a[maxn+100];

 6 int bb[maxn+100]; //存储单个元素的逆序数

 7 int main()

 8 {

 9     int n,i,j,tol;

10     while(scanf("%d",&n)!=EOF)

11     {

12         memset(bb,0,sizeof(bb));

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

14         {

15           scanf("%d",a+i);

16           for(j=i-1;j>=0;j--)

17           {

18               if(a[i]>a[j]&&bb[j]==0) break;

19               if(a[i]<a[j])bb[i]++;

20           }

21         }

22         tol=0;

23         for(i=0;i<n;i++)  //求出逆序数

24              tol+=bb[i];

25              int res=tol;

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

27         {

28                 tol+=n-2*a[i]-1 ;

29                 if(res>tol)

30                     res=tol;

31         }

32         printf("%d\n",res);

33     }

34 

35     return 0;

36 }

 

 

运用递归调用版的归并排序

比如 5 4 3 2 1 《5 ，4》，《3 ，2》  --》+ 2

 4 5 2 3 1

4 5 2 3  ---》 2 +2=4；

2 3 4 5 1 --》 4

10

运用这个原理便可以得到结果，代码如下：

 1 #include<string.h>

 2 #include<stdlib.h>

 3 #include<stdio.h>

 4 #define maxn 5000

 5 int aa[maxn+100];

 6 int bb[maxn+100];

 7 int nn,tol=0;

 8 void mergec(int low ,int mid ,int hight )

 9 {

10     int i,j,k;

11    int *cc = (int *)malloc(sizeof(int)*(hight-low+3));

12      i=low;

13      j=mid;

14      k=0;

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

16     {

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

18         {

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

20           tol+=mid-i;

21         }

22        else

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

24     }

25     for( ; i<mid ;i++)

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

27     for( ; j<hight ; j++)

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

29     k=0;

30     for(i=low;i<hight;i++)

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

32       free( cc );

33 }

34 /*用递归求解归并排序无法求逆序数*/

35 void merge_sort(int st,int en)

36 {

37     int mid;

38     if(st+1<en)

39     {

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

41         merge_sort(st,mid);

42         merge_sort(mid,en);

43         mergec(st,mid,en);

44     }

45 }

46 int main()

47 {

48     int  i,res;

49  // freopen("test.in","r",stdin);

50     while(scanf("%d",&nn)!=EOF)

51     {

52           tol=0;

53         for(i=0;i<nn;i++){

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

55              bb[i]=aa[i];

56          }

57          merge_sort(0,nn);

58           res=tol;

59        //printf("tol=%d\n",res);

60          for(i=0;i<nn-1;i++)

61          {

62              tol+=nn-2*bb[i]-1;

63              if(res>tol) res=tol;

64          }

65         printf("%d\n",res);

66     }

67     return 0;

68 }

接下来是非递归调用....版的归并排序

 

 

 1 #include<string.h>

 2 #include<stdlib.h>

 3 #include<stdio.h>

 4 #define maxn 5000

 5 int aa[maxn+100];

 6 int bb[maxn+100];

 7 int nn,tol=0;

 8 void mergec(int low ,int mid ,int hight )

 9 {

10     int i,j,k;

11    int *cc = (int *)malloc(sizeof(int)*(hight-low+3));

12      i=low;

13      j=mid;

14      k=0;

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

16     {

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

18         {

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

20           tol+=mid-i;

21         }

22        else

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

24     }

25     for( ; i<mid ;i++)

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

27     for( ; j<hight ; j++)

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

29     k=0;

30     for(i=low;i<hight;i++)

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

32       free( cc );

33 }

34 

35 /*----------------------华丽丽的分割线--------------------------------*/

36 void merge_sort( int st , int en )

37 {

38     int s,t,i;

39     t=1;

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

41     {

42         s=t;

43         t=s*2;    //表示两个s的长度

44         i=st;

45         while(i+t<=en){

46             mergec(i,i+s,i+t);

47             i+=t;

48         }

49      if(i+s<en)

50          mergec(i,i+s,en);

51     }

52     if(s<en-st)

53          mergec(st,st+s,en);

54 }

55 int main()

56 {

57     int  i,res;

58 //  freopen("test.in","r",stdin);

59     while(scanf("%d",&nn)!=EOF)

60     {

61           tol=0;

62         for(i=0;i<nn;i++){

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

64              bb[i]=aa[i];

65          }

66          merge_sort(0,nn);

67           res=tol;

68        //printf("tol=%d\n",res);

69          for(i=0;i<nn-1;i++)

70          {

71              tol+=nn-2*bb[i]-1;

72              if(res>tol) res=tol;

73          }

74         printf("%d\n",res);

75     }

76     return 0;

77 }

 

 用树状数组...

代码：

 1 /*

 2 用树状数组求逆序数

 3 */

 4 #include<stdio.h>

 5 #include<string.h>

 6 #include<stdlib.h>

 7 #define maxn 5000

 8 int aa[maxn+100];

 9 int bb[maxn+100];

10 int nn;

11 int lowbit(int k)

12 {

13    return k&(-k);

14 }

15 void ope(int x)

16 {

17     while(x<=nn)

18     {

19       aa[x]++;

20       x+=lowbit(x);

21     }

22 }

23 int sum(int x)

24 {

25     int ans=0;

26     while(x>0)

27     {

28         ans+=aa[x];

29         x-=lowbit(x);

30     }

31     return ans;

32 }

33 int main()

34 {

35 

36     int i,res,ans;

37     //freopen("test.in","r",stdin);

38     while(scanf("%d",&nn)!=EOF)

39     {

40        memset(aa,0,sizeof(aa));

41        res=0;

42        for(i=0;i<nn;i++)

43        {

44          scanf("%d",&bb[i]);

45          res+=sum(nn)-sum(bb[i]+1);

46          ope(bb[i]+1);

47        }

48        ans=res;

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

50        {

51            res+=nn-1-2*bb[i];

52            if(ans>res)

53                   ans=res;

54        }

55        printf("%d\n",ans);

56    }

57   return 0;

58 }