题目传送门
Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25116 Accepted Submission(s): 14827
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
Sample Output
16
Author
CHEN, Gaoli
Source
Recommend
题意：给你一个环状数列，比如1234，转一下成2341等等，问你在每次转化的时候，
形成的数列最小的逆序数是多少？
题解：先用O（logn）的时间求出原始数列的逆序数，然后我们从上面的转移中可以发现，
每次形成的新数列都是原数列头一个数被转移到最末的位置，即每次要从之前的逆序数
先-a[i]（失去开头数的逆序数），再+n-1-a[i]（在末尾新增的逆序数）；最后一直枚举
下来边求最小值即可
代码：
#include<iostream>
#include<string.h>
#include<algorithm>
#include<stdio.h>
#include<queue>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long long ll;
typedef pair<int,int> PII;
#define mod 1000000007
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
//head
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const int maxn=5005;
int sum[maxn<<2];
void PushUP(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt)
{
sum[rt]=0;
if(l==r) return ;
int m=(l+r)>>1;
build(lson);
build(rson);
PushUP(rt);
}
void update(int p,int l,int r,int rt)
{
if(l==r){
sum[rt]++;
return ;
}
int m=(l+r)>>1;
if(p<=m){
update(p,lson);
}
else update(p,rson);
PushUP(rt);
}
int query(int L,int R,int l,int r,int rt)
{
if(L<=l&&r<=R)
return sum[rt];
int ret=0;
int m=(l+r)>>1;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson);
return ret;
}
int main()
{
int n;
int a[maxn];
while(~scanf("%d",&n))
{
build(0,n-1,1);
int ans=0;
for(int i=0;i<n;i++){
scanf("%d",&a[i]);
ans+=query(a[i],n-1,0,n-1,1);
update(a[i],0,n-1,1);
}
int cnt=ans;
for(int i=0;i<n;i++)
{
ans+=n-a[i]-a[i]-1;
cnt=min(ans,cnt);
}
printf("%d
",cnt);
}
return 0;
}