Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7428 Accepted Submission(s): 4560
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.
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
1 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
CHEN, Gaoli
Source
Recommend
Ignatius.L
#include <iostream>
#include <cstdio>
#include <cstring>
#define cls(x) memset((x),0,sizeof((x)));
using namespace std;
int a[5100],x[5100],n,cnt,ans,sum;
int main()
{
while(scanf("%d",&n)!=EOF)
{
sum=0;cls(a);cls(x);
for(int i=0;i<n;i++)
{
scanf("%d",&a);cnt=0;
for(int j=0;j<i;j++)
{
if(a[j]>a)
cnt++;
}
sum+=cnt;
}
for(int i=0;i<n;i++)
{
cnt=0;
for(int j=0;j<n;j++)
{
if(a[j]<a&&i!=j)
cnt++;
}
x=cnt;
}
ans=sum;
for(int i=0;i<n;i++)
{
sum=sum-2*x+n-1;
ans=min(sum,ans);
}
printf("%d
",ans);
}
return 0;
}