hdu1394（树状数组）最小逆序对数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 22426    Accepted Submission(s): 13351


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

 

4000

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#include<iostream>
#include<cstdio>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<list>
#include<set>
#include<iomanip>
#include<cstring>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<cassert>
#include<sstream>
#include<algorithm>
using namespace std;
const int MAXN = 10005;
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define PI numcos(-1.0)
typedef long long ll;

int c[MAXN],num[MAXN],n;

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

void add(int i,int value)
{
while(i<=n)
{
c[i]+=value;
i+=lowbit(i);
}
}

int sum(int i)
{
int sum=0;
while(i>0)
{
sum+=c[i];
i-=lowbit(i);
}
return sum;
}

int main()
{
while(~scanf("%d",&n))
{
memset(c,0,sizeof(c));
for(int i=0; i<n; i++)
scanf("%d",&num[i]);
int ans=0,tp;
for(int i=0; i<n; i++)//n-num[i]表示比num[i]大的数有多少个
{
ans+=sum(n-num[i]);//当前num[i]的逆序对数有多少
add(n-num[i],1);//
}
tp=ans;
for(int i=0; i<n-1; i++)
{
add(n-num[i],-1);//把第一个数字从所放位置拿掉
tp=tp+sum(n-num[i])-num[i];
//原逆序对数+第一个数字放在最后一位上时能产生的逆序对数-第一位以后有多少个比第一位小的数字数目
add(n-num[i],1);//数字放回
if(ans>tp) ans=tp;
}
printf("%d\n",ans);
}
return 0;
}