Minimum Inversion Number HDU - 1394

Time limit
1000 ms
Memory limit
32768 kB

     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

最好用线段树，不过暴力跑也可以过。
由于懒的写线段树所以我用的是暴力的方法，这里也只有暴力的代码。

思路：
以1 3 6 9 0 8 5 7 4 2为例，这十个数中比1小的有一个，比1大的有8个，
所以当1换到最后时此时关于1的逆序数从一个变成了八个，3类似，依次往下推直到以2为首，则过程中的最小值即为结果。

代码：

#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

int board[5005][2];//分别记录每个点的值和每个点在数列中比其小的点的数量。

int main(){
    int N;
    while(scanf("%d",&N)!=EOF){
        int sum = 0;//数列初始时的总逆序数
        for(int i=0 ; i<N ; i++)scanf("%d",&board[i][0]);

        for(int i=0 ; i<N ; i++){
            int mid = 0;
            for(int j=0 ; j<i ; j++){
                if(board[j][0]<board[i][0])mid++;
            }
            board[i][1] = mid;
        }

        for(int i=0 ; i<N ; i++){
            int mid = 0;
            for(int j=i+1 ; j<N ; j++){
                if(board[j][0]<board[i][0])mid++;
            }
            board[i][1] += mid;
            sum += mid;
        }

        int re = sum;
        for(int i=0 ; i<N ; i++){
            sum = sum-board[i][1] + N-board[i][1]-1;
            re = min(re,sum);
        }
        printf("%d\n",re);
    }
    return 0;
}