F - Minimum Inversion Number-优快云博客

本文介绍了一种算法，用于解决给定序列通过循环移位获得的所有序列中，逆序对数量最少的问题。该文提供了两种不同的实现方案，一种使用了直接计算的方法，另一种则利用了归并排序的思想。

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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

/*#include <stdio.h>
#include <string.h>
#include<iostream>
using namespace std;
int main()
{
int n;
int a[5005];
while(~scanf("%d",&n))
{
long long num = 0;
for(int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
if(i != 0)
{
for(int j = i-1; j >= 0; j--)
{
if(a[j] > a[i])
num++;
}
}
}
//printf("%d\n",num);
long long ans = num;
for(int i = 0; i < n; i++)
{
num = num + n-1-2*a[i];
if(ans > num)
ans = num;
}
printf("%lld\n",ans);

}
return 0;
}*/
#include <stdio.h>
#include <algorithm>

using namespace std;

int a[5005];
struct Node{
int l,r,num;
}tree[50000];

void Build(int n,int x,int y){
tree[n].l = x;
tree[n].r = y;
tree[n].num = 0;
if(x == y){
return;
}
int mid = (x + y) / 2;
Build(2*n,x,mid);
Build(2*n+1,mid+1,y);
}

void Modify(int n,int x){
int l = tree[n].l;
int r = tree[n].r;
int mid = (l + r) / 2;
if(x == l && x == r){
tree[n].num = 1;
return;
}
if(x <= mid) Modify(2*n,x);
else Modify(2*n+1,x);
tree[n].num = tree[2*n].num + tree[2*n+1].num;
}

int Query(int n,int x,int y){
int l = tree[n].l;
int r = tree[n].r;
int mid = (l + r) / 2;
int ans = 0;;
if(x == l && y == r)
return tree[n].num;
if(x <= mid) ans += Query(2*n,x,min(mid,y));
if(y > mid) ans += Query(2*n+1,max(mid+1,x),y);
return ans;
}
int main(){
int n,sum,ans;
int i,j;

while(scanf("%d",&n) != EOF){
sum = 0;
Build(1,0,n);
for(i = 1;i <= n;i++){
scanf("%d",&a[i]);
Modify(1,a[i]);
sum += Query(1,a[i]+1,n);
}
ans = sum;
for(i = 1;i < n;i++){
sum = sum + (n - 1 - a[i]) - a[i];
if(sum < ans)
ans = sum;
}
printf("%d\n",ans);
}
}
归并排序：
#include<cstdio>
#include<iostream>
using namespace std;
const int maxn=5005;
int num1[maxn],num2[maxn],temp[maxn];
int sum;
void Merge(int l,int mid,int r){
int p=0;
int i=l,j=mid+1;
while(i<=mid&&j<=r){
if(num1[i]>num1[j]){
sum+=mid-i+1;
temp[p++]=num1[j++];
}
else temp[p++]=num1[i++];
}
while(i<=mid)temp[p++]=num1[i++];
while(j<=r)temp[p++]=num1[j++];
for(i=0;i<p;i++)
num1[l+i]=temp[i];
}
void MergeSort(int l,int r){
if(l<r){
int mid=(l+r)>>1;
MergeSort(l,mid);
MergeSort(mid+1,r);
Merge(l,mid,r);
}
}
int main()
{
int n;
while(~scanf("%d",&n)){
for(int i=0;i<n;i++){
scanf("%d",&num1[i]);
num2[i]=num1[i];
}
sum=0;
MergeSort(0,n-1);
int ans=sum;
for(int i=0;i<n-1;i++){
sum+=n-num2[i]-num2[i]-1;
ans=ans<sum?ans:sum;
}
printf("%d\n",ans);
}
return 0;
}