HDU 1394Minimum Inversion Number 数状数组 逆序对数量和

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18543    Accepted Submission(s): 11246

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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1394

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题意：求出对(ai,aj)(i<j,ai>aj)的全部数量。即求ai右边比ai小的数的个数和。每次变换a的序列，求出最小的逆序对数量和。

思路： 因为树状数组的最基本功能就是求比某点 x 小的点的个数。所以逆向存储ai。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN=1e6+100,INF=1e9+100;
int a[MAXN],c[MAXN],ans[MAXN];
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int val)
{
    for(i; i<=MAXN; i+=lowbit(i))
        c[i]+=val;
}
int sum(int i)
{
    int s=0;
    for(i; i>0; i-=lowbit(i))
        s+=c[i];
    return s;
}
int main()
{
    int i,j,t,n;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=n; i>=1; i--)
            scanf("%d",&a[i]);
        memset(c,0,sizeof(c));
        memset(ans,0,sizeof(ans));
        int cou=0;
        for(i=1; i<=n; i++)
        {
            ans[i]=sum(a[i]+1);
            cou+=ans[i];
            add(a[i]+1,1);
        }
        int Min=cou;
        for(i=n; i>1; i--)
        {
            for(j=1; j<=n; j++)
            {
                if(j==i) continue;
                if(a[i]<a[j])
                {
                    cou++;
                    ans[j]++;
                }
            }
            cou-=ans[i];
            ans[i]=0;
            if(cou<Min) Min=cou;
        }
        cout<<Min<<endl;
    }
    return 0;
}