Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24289 Accepted Submission(s): 14383
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
分析:
可利用归并排序求出第一次的逆序数和,在用下面的公式求解最小的逆序数和;
如果每次都进行如上操作,那么最多进行n-1次后会与原序列重合;
这个序列的元素是从0---(n-1),当某个元素位于首位置时,其本身的数值就是后面的逆序个数;
每次将它移至末尾,原来的逆序即变成了正序,正序变逆;由此可得 sum=sum-a[i]+(n-a[i]-1);
#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>
#include <algorithm>
using namespace std;
#define Min(a,b)(a<b?a:b)
const int maxn=5050;
int a[maxn],b[maxn],c[maxn];
int sum;
void Merge(int begin,int mid,int end){
int i=begin,j=mid+1,pos=begin;
while(i<=mid && j<=end){
if(a[i]<=a[j]){
b[pos++]=a[i++];
}else{
b[pos++]=a[j++];
sum+=mid-i+1;//求逆序数
}
}
while(i<=mid) b[pos++]=a[i++];
while(j<=end) b[pos++]=a[j++];
for(int i=begin,j=begin;i<=end;i++,j++)
a[i]=b[j];
}
void Sort(int begin,int end){
if(begin<end){
int mid=(begin+end)/2;
Sort(begin,mid);
Sort(mid+1,end);
Merge(begin,mid,end);
}
}
int main(){
int n;
while(~scanf("%d",&n))
{
int ans;
sum=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
c[i]=a[i];
}
Sort(1,n);
ans=sum;
for(int i=1;i<=n;i++)
{
sum=sum-c[i]+(n-c[i]-1);
ans=Min(ans,sum);
}
printf("%d\n",ans);
}
return 0;
}