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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
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 }
HDU 1394 - Minimum Inversion Number
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原文地址：http://www.cnblogs.com/nicetomeetu/p/5165048.html
