hdu1394(点树在逆序数中的应用)

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K(Java/Others)
Total Submission(s): 5005 Accepted Submission(s):3063

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 thefirst m >= 0 numbers to the end of the seqence, wewill obtain another sequence. There are totally n such sequences asthe 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 inversionnumber out of the above sequences.

 

Input

The input consists of a number of test cases. Each caseconsists of two lines: the first line contains a positive integer n(n <= 5000); the next line contains a permutation ofthe n integers from 0 to n-1.

 

Output

For each case, output the minimum inversion number on a singleline.

 

Sample Input

10 1 3 6 90 8 5 7 4 2

 

Sample Output

16

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

Recommend

Ignatius.L

//左移序列之后得到新的序列的逆序数的个数就等与当前总的个数减左边的数的逆序数a[i]，然后加上新增的逆序数n-a[i]-1；
#include
#include
#include
using namespace std;
const int N = 5050;
struct node
{
 int l,r;
 int num; //表示该区间已经出现节点的个数
}tree[N*4];
void bulid(int rt ,int l,int r)//建树
{
 tree[rt].l=l;
 tree[rt].r=r;
 tree[rt].num=0;//********
 if(l==r)
  return;
 int mid = (l+r)/2;
 bulid(2*rt,l,mid);//创建左子树
 bulid(2*rt+1,mid+1,r);//创建右子树
}
int search(int rt, int l,int r)
{
 if(l>r)//剪枝
  return 0;
 if(l==tree[rt].l&&r== tree[rt].r)
  return tree[rt].num;
 int mid = (tree[rt].l+tree[rt].r)/2;
 if(r<=mid)//建立左子树
  return search(2*rt,l,r);
 else if(l>mid)//建立右子树
  returnsearch(2*rt+1,l,r);
 else
  returnsearch(2*rt,l,mid)+search(2*rt+1,mid+1,r);
}
void update(int rt, int x)
{
 tree[rt].num++;
 if(tree[rt].l==tree[rt].r)
  return;
 int mid = (tree[rt].l+tree[rt].r)/2;
 if(x<=mid)
  update(2*rt,x);
 else update(2*rt+1,x);
}
int main ()
{
 int a[N];
 int n;
 int i, min;
 while(scanf("%d",&n)!=EOF)
 {
  
  bulid(1,0,n-1);
  int sum = 0;
  for(i=0;i
     scanf("%d",&a[i]);
  for(i=0;i
  {
    sum+=search(1,a[i]+1,n-1);//查找在a[i]之前出现且比a[i]大的数
   cout<< sum<< endl;
   update(1,a[i]);
  }
  min =sum;
  for(i=0;i
  {
   sum +=(n-a[i]-1)-a[i];//关键
   if(min>sum)
    min=sum;
  }
  printf("%d\n",min);
 }
 return 0;
}