hdu 1394 Minimum Inversion Number 归并求逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12107 Accepted Submission(s): 7388



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

	昨天做了几道逆序数的题，全用的树状数组解决的。晚上睡觉的时候发现，树状数组求逆序数，还是有很多麻烦的问题的，1，当序列的的离散程度比较大的大时候，，直接用树状数组，，会MLE，需要离散化序列，虽然不是太难，可模块一多，难免会出现一些比较隐晦的bug，在比赛的时候最忌讳这个了。	2，当序列中有负数的时候，，可能我们就得重新选定坐标原点了，这个原点的坐标得根据数据的范围来选，如果题目说明的不清楚的时候，，就很坑了。	所以我决定再复习一下归并求逆序数，毕竟这个没什么限制，虽然速度不太怎么满意，但好在实用性比较强。	这道题就是给定一个序列，a1, a2, ..., an-1, an (where m = 0 - the initial seqence)让你把按照这些顺序的序列的逆序数全求出来：	a2, a3, ..., an, a1 (where m = 1)
	a3, a4, ..., an, a1, a2 (where m = 2)
	...
	an, a1, a2, ..., an-1 (where m = n-1)
然后把最小的输出来。没什么好说的，直接模板，下面是代码：
#include <stdio.h>
#define MAX 10000
#define INF 100000000
int b[MAX];

int merge(int a[] , int start , int mid , int end)
{
	int i = start , j = mid+1 , k = start , count = 0;
	while(i<=mid && j<=end)
	{
		if(a[i]<=a[j])
		{
			b[k++] = a[i++];
		}
		else
		{
			b[k++] = a[j++];
			count += mid-i+1;
		}
	}
	while(i<=mid)
	{
		b[k++] = a[i++] ;
	}
	while(j<=end)
	{
		b[k++] = a[j++] ;
	}
	for(i = start ; i <= end ; ++i)
	{
		a[i] = b[i] ;
	}
	return count ;
}

int mergeSort(int a[],int start , int end)
{
	int sum = 0 ;
	if(start == end)
	{
		return 0;
	}
	int mid = (start+end)>>1 ;
	sum +=mergeSort(a,start,mid) ;
	sum +=mergeSort(a,mid+1,end) ;
	sum +=merge(a,start,mid,end) ;
	return sum ;
}

int min(int a , int b)
{
	return a>b?b:a ;
}
int main()
{
	int n ;
	while(~scanf("%d",&n))
	{
		int a[MAX],temp[MAX];
		for(int i = 0 ; i < n ; ++i)
		{
			scanf("%d",&a[i]);
			temp[i] = a[i] ;
		}
		int sum = INF ,ans;
		sum = mergeSort(a,0,n-1) ;
		ans = sum ;
		for(int i = 0 ; i < n-1 ; ++i)
		{
			sum += -temp[i]+ n-temp[i]-1;
			ans = min(sum,ans) ;
		}
		
		printf("%d\n",ans) ; 
	}
	
	return 0;
}

AC状态：