HDU1394_Minimum Inversion Number_线段树求逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19452    Accepted Submission(s): 11701

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
  
  
  
  


  
  

大致题意：

对于给出的数组，每次把第一个元素挪到最后，求此过程中最小的逆序数。

大体思路：

可以用归并排序求。这里说用线段树的解法。

处理初始数组：对于每一个数，查找线段树中已有的比他大的数的数量。之后把这个数加入线段树。

每一次移动：假设这次移动的是m，则他后面比m小的有 m-1 个，比m大的有 n-m个，所以新的逆序数等于 原来的逆序数 - （m-1） + n - m.

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

#define _max 5010

int Seg [_max<<2];
int A [_max] ;
int n ;

int Query (int p, int l, int r, int x, int y)
{
	if(x<=l && y>=r) return Seg[p];
	int mid = (l+r)>>1,ans = 0;
	if(x<=mid) ans += Query(2*p, l, mid, x, y);
	if(y> mid) ans += Query(2*p+1, mid+1, r, x, y);
	return ans;
}

void Update (int p, int l, int r, int x)
{
	if(l==r){
		Seg[p]++;
		return;
	}
	int mid = (l+r)>>1;
	if(x<=mid) Update(2*p, l, mid, x);
	else Update(2*p+1, mid+1, r, x);
	Seg[p] = Seg[2*p] + Seg[2*p+1];
}

int main ()
{
	//freopen("in.txt","r",stdin);

	while(scanf("%d",&n) != EOF){

		memset(Seg,0,sizeof(Seg));

		for( int i=1; i<=n; i++){
			scanf("%d",A+i);
			A[i]++;
		}

		int ans = 0 ;
		for(int i=1; i<=n; i++){
			ans += Query(1, 1, n, A[i], n);
			Update(1, 1, n, A[i]);
		}

		int Min = ans;
		for(int i=1; i<=n; i++){
			ans = ans - A[i] + 1 + n - A[i] ;
			if(ans<Min) Min = ans ;
		}

		printf("%d\n",Min);

	}

	return 0 ;

}