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
就先求第一个顺序的逆序数 。
用线段树求逆序数的方法就是下标代表那个值 0，1代表有没有出现 这样每次查询后面的值之和就是当前值对应的逆序数啦
但是要注意因为从0开始所以建树要从0开始
求玩第一个顺序别的也就好求了
现在把A1从队首挪到队尾 当A1在队首时 后面得数有多少个比它小呢？a1个 挪到队尾A1的逆序数是多少？n-a1-1
#include<bits/stdc++.h>
using namespace std;
const int maxn=5005;
struct node
{
int l,r,v;
}tree[maxn];
int a[maxn];
void build(int i,int l,int r)
{
tree[i].l=l;
tree[i].r=r;
if(l==r)
{
tree[i].v=0;
return ;
}
int m=(l+r)/2;
build(i*2,l,m);
build(i*2+1,m+1,r);
tree[i].v=tree[i*2].v+tree[i*2+1].v;
}
void update(int i,int x,int v)
{
if(tree[i].l==tree[i].r)
{
tree[i].v=1;
//cout<<tree[i].l<<endl;
return ;
}
int m=(tree[i].l+tree[i].r)/2;
if(x>m)update(i*2+1,x,v);
else update(i*2,x,v);
tree[i].v=tree[i*2].v+tree[i*2+1].v;
}
int query(int i,int x,int y)
{
int sum=0;
// cout<<x<<" "<<tree[i].l<<endl;
if(x<=tree[i].l&&tree[i].r<=y)
{
//cout<<"nsdkfjsd"<<endl;
//cout<<tree[i].l<<" "<<tree[i].r<<endl;
sum+=tree[i].v;
return sum;
}
int m=(tree[i].l+tree[i].r)/2;
if(x<=m)sum+=query(i*2,x,y);
if(y>m)sum+=query(i*2+1,x,y);
return sum;
}
int main()
{
int n;
while(cin>>n)
{
build(1,0,n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int sum=0;
for(int i=1;i<=n;i++)
{
sum+=query(1,a[i],n);
//cout<<i<<" "<<sum<<endl;
update(1,a[i],1);
}
int mmin=sum;
for(int i=1;i<=n;i++)
{
sum-=a[i];
sum+=n-a[i]-1;
mmin=min(sum,mmin);
}
printf("%d\n",mmin);
}
return 0;
}
