hdu 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7582    Accepted Submission(s): 4649

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

 

本題較水，我用暴力過了，就不想打線段樹！

AC代碼（暴力實現）

#include<stdio.h> 
int a[5003]; 
int main() 
{     
	int n,i,j,sum;   
	while(scanf("%d",&n)!=EOF)  
	{    
		sum=0;  
        for(i=1;i<=n;i++)   
		{             
			scanf("%d",a+i);    
			for(j=1;j<i;j++)   
				if(a[j]>a[i])      
					sum++;      
		}        
		  int ans=sum;      
		 
		  for(i=1;i<=n;i++)   
		  {     
			  sum+=(-a[i]+n-a[i]-1); 
			 //交換倆個不相鄰的數a，b（a在前，b在後），逆序數+=兩者間大於a的個數-兩者間大於b的個數！！！
			  if(sum<ans)           
				  ans=sum;   
          }        
		  printf("%d\n",ans); 
    } 
	return 0;
}