HDU 1394 - Minimum Inversion Number

Posted 2020-06-09

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

 

大致题意：

　　一个由0..n-1组成的序列，每次可以把队首的元素移到队尾；
      求形成的n个序列中最小逆序对数目；

解题思路：

　　暴力法：

　　　　当吧 a[0] 移到队尾后，会减少 a[0] 个逆序对 同时会增加  (n-1)-a[0] 个逆序对；

　　　　那么只要求出初始的逆序对数好了，剩下的可以推导得出；

 

 1 #include <cstdio>
 2 using namespace std;
 3 int main(){
 4     int n,a[5000+5],t;
 5     while(~scanf("%d",&n)){
 6         for(int i=0;i<n;i++) scanf("%d",&a[i]);
 7         t=0;
 8         for(int i=0;i<n;i++)
 9             for(int j=i+1;j<n;j++)
10                 if(a[i]>a[j]) t++;
11         int min=t;
12         for(int i=0;i<n;i++){
13             t=t-a[i]+(n-1)-a[i];
14             if(t<min) min=t;
15         }
16         printf("%d\n",min);
17     }
18 }