hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16188 Accepted Submission(s): 9845

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

1 3 6 9 0 8 5 7 4 2

Sample Output

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

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复制了大神的博客：

弄了半天才弄懂题目的意思，就是求最小的逆序数，在此贴下逆序数的概念

在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。逆序数为偶数的排列称为偶排列；逆序数为奇数的排列称为奇排列。如2431中，21，43，41，31是逆序，逆序数是4，为偶排列。

也是就说，对于n个不同的元素，先规定各元素之间有一个标准次序（例如n个不同的自然数，可规定从小到大为标准次序），于是在这n个元素的任一排列中，当某两个元素的先后次序与标准次序不同时，就说有1个逆序。一个排列中所有逆序总数叫做这个排列的逆序数。

题目的意思就好比给出一个序列

如：0 3 4 1 2

设逆序数初始n = 0；

由于0后面没有比它小的，n = 0

3后面有1,2 n = 2

4后面有1,2,n = 2+2 = 4；

所以该序列逆序数为 4

其根据题意移动产生的序列有

3 4 1 2 0 逆序数：8

4 1 2 0 3 逆序数：6

1 2 0 3 4 逆序数：2

2 0 3 4 1 逆序数：4

所以最小逆序数为2

如果是0到n的排列，那么如果把第一个数放到最后，对于这个数列，逆序数是减少a[i],而增加n-1-a[i]的。然后就解决了。

附上代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #define M 5005
 5 using namespace std;
 6 
 7 struct node
 8 {
 9     int l,r,n;
10 } ss[M*3];
11 int n;
12 
13 void build(int l,int r,int k)
14 {
15     ss[k].l=l;
16     ss[k].r=r;
17     ss[k].n=0;
18     if(l==r) return;
19     int mid=(ss[k].l+ss[k].r)/2;
20     build(l,mid,k*2);
21     build(mid+1,r,k*2+1);
22 }
23 
24 void insert(int x,int k)
25 {
26     if(ss[k].l==x&&ss[k].r==x)
27     {
28         ss[k].n=1;
29         return;
30     }
31     int mid=(ss[k].l+ss[k].r)/2;
32     if(x<=mid) insert(x,k*2);
33     else insert(x,k*2+1);
34     ss[k].n=ss[k*2].n+ss[k*2+1].n;
35 }
36 
37 int search(int x,int k)
38 {
39     if(ss[k].l>=x&&ss[k].r<n) return ss[k].n;
40     else
41     {
42         int sum1=0,sum2=0;
43         int mid=(ss[k].l+ss[k].r)/2;
44         if(x<=mid) sum1=search(x,k*2);
45         if(n-1>mid) sum2=search(x,k*2+1);
46         return sum1+sum2;
47     }
48 }
49 
50 int main()
51 {
52     int i,j,m;
53     int a[M];
54     while(~scanf("%d",&n))
55     {
56         build(0,n-1,1);
57         int sum=0;
58         for(i=0; i<n; i++)
59         {
60             scanf("%d",&a[i]);
61             sum+=search(a[i]+1,1);
62             insert(a[i],1);
63         }
64         int minx=sum;
65         for(i=0; i<n; i++)
66         {
67             sum=sum+n-2*a[i]-1;
68             if(sum<minx) minx=sum;
69         }
70         printf("%d
",minx);
71 
72     }
73     return 0;
74 }