Minimum Inversion Number
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
Sample Output
Author
CHEN, Gaoli
Source
ZOJ Monthly, January 2003
sum记录初始的逆序对数
之后对sum逆序循环操作 循环到i表示处理将第i个数从数列末尾调到开头的情况
总共有n个数,对于数a[i]在数列末尾时前面比他大的个数为n-a[i]-1,比他小的个数为a[i]
当把a[i]调到开头时原先比他大的(n-a[i]-1)个所构成的逆序对变成正序 故sum-n+a[i]+1
原先比a[i]小的a[i]个正序对变成逆序对 故(sum-n+a[i]+1)+a[i]
循环一遍 取最小的sum 就是答案
#include<cstdio>
#include<cstring>
using namespace std;
#define lowbit(x) x&(-x)
#define N 5000+5
#define INF 1000000000
int cnt[N];
int add(int n)
{
while(n<=N)
{
cnt[n]++;
n+=lowbit(n);
}
return 0;
}
int query(int n)
{
int sum=0;
while(n>0)
{
sum+=cnt[n];
n-=lowbit(n);
}
return sum;
}
int main()
{
int a[N],n,sum,i,min;
while(scanf("%d",&n)!=EOF)
{
min=INF;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sum=0;
memset(cnt,0,sizeof(cnt));
for(i=1;i<=n;i++)
{
add(a[i]+1);
sum+=i-query(a[i]+1);
}
for(i=n;i>1;i--)
{
sum=sum-n+1+2*a[i];
if(sum<min)
min=sum;
}
printf("%d\n",min);
}
return 0;
}