hdu 1394 Minimum Inversion Number (树状数组 逆序对)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15247    Accepted Submission(s): 9311

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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虽然在学线段树。。。但是感觉美必要为了用线段树而用线段树。。。

我知道树状数组的东西都可以用线段树来解决。。。。

不过还是觉得树状数组。。。实在是优美。。。所以这题还是用树状数组搞得。。

 

这题是问一个长度为n的循环数组中，逆序对最少的个数。。。

我们可以先用树状数组求出初始的数列的逆序对。。。

然后其他的可以通过递推得到。。。。

当a[i]从处于位置1而被放到最后的时候。。。

cnt = cnt -a[i]+n-a[i]+1;

然后取所有cnt的最大值就行。

/*************************************************************************
    > File Name: code/hud/1394.cpp
    > Author: 111qqz
    > Email: rkz2013@126.com 
    > Created Time: 2015年10月28日 星期三 21时16分47秒
 ************************************************************************/

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<map>
#include<set>
#include<queue>
#include<vector>
#include<stack>
#include<cctype>
                 
#define yn hez111qqz
#define j1 cute111qqz
#define ms(a,x) memset(a,x,sizeof(a))
using namespace std;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
typedef long long LL;
typedef double DB;
const int inf = 0x3f3f3f3f;
const int N=5E3+7;
int c[N];
int n,a[N];

int lowbit( int x)
{
    return x&(-x);
}


void update ( int x,int delta)
{
   // cout<<"a?"<<endl;
    for ( int i = x ; i <= n ; i = i + lowbit(i)) c[i] = c[i] + delta;
}
int Sum( int x)
{
 //   cout<<"b?"<<endl;
    int res = 0 ;
    for ( int i = x ;i >= 1; i = i - lowbit(i))
    {
    res = res + c[i];
    }
    return res;
}
int main()
{
  #ifndef  ONLINE_JUDGE 
   freopen("in.txt","r",stdin);
  #endif
   while (scanf("%d",&n)!=EOF)
    {
    ms(c,0);
    for ( int i = 1 ; i <= n ; i++)
    {
        scanf("%d",&a[i]);
        a[i]++;
    }
    int cnt = 0 ;
    int ans = inf;
 //   puts("mmmmmmmiao?");
    for ( int i = 1  ; i <= n ; i++)
    {
    update(a[i],1);
    cnt = cnt + i - Sum(a[i]);
//    puts("whats wrong?");
    }
  //  printf("cnt::%d
",cnt);
    if (cnt<ans) ans = cnt;
    for ( int i = 1 ; i <= n ; i++)
    {
    cnt =cnt -a[i]+n-a[i]+1;
    if (cnt<ans&&cnt>0) ans = cnt;
//    printf("cnt:%d
",cnt);
    }
    printf("%d
",ans);
    }
    
  
   
 #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}