• 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
     
           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
    ",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
    ",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
    ",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
    ",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
    ",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
    ",ans);
    56    }
    57   return 0;
    58 }
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  • 原文地址:https://www.cnblogs.com/gongxijun/p/3650454.html
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