Minimum Inversion Number（逆序数）（技巧or树状数组or线段树）

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

转载来之：https://wenku.baidu.com/view/6e02b7492e3f5727a5e9623f.html

代码如下：

#include<stdio.h>
int a[5003];
int main()
{
    int n,i,j,sum;
    while(scanf("%d",&n)!=EOF)
    {
        sum=0;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            for(j=1;j<i;j++)
            if(a[j]>a[i])
            sum++;
        }
        int ans=sum;
        for(i=1;i<=n;i++)
        {
            sum=sum+(-a[i]+n-1-a[i]);
            if(sum<ans)
            ans=sum;
        }
        printf("%d\n",ans);
    }    
 return 0;
}

就是有前面多少个比他大的 就是逆序       所有加起来就是逆序数

接下来是树状数组 转载来自http://www.cnblogs.com/kuangbin/archive/2012/08/15/2639385.html
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int MAXN=5050;
int c[MAXN];
int n;
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    while(i<=n)
    {
        c[i]+=val;
        i+=lowbit(i);
    }
}
int sum(int i)
{
    int s=0;
    while(i>0)
    {
        s+=c[i];
        i-=lowbit(i);
    }
    return s;
}
int a[MAXN];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        int ans=0;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans+=sum(n)-sum(a[i]);
            add(a[i],1);
        }
        int Min=ans;
        for(int i=1;i<=n;i++)
        {
            ans+=n-a[i]-(a[i]-1);
            if(ans<Min)Min=ans;
        }
        printf("%d\n",Min);
    }
    return 0;
}