HDU 1394（最小逆序数）【也可用线段树求解】

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 753    Accepted Submission(s): 402

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

101 3 6 9 0 8 5 7 4 2

 

 

Sample Output

16
先说一下逆序数的概念：
在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，那末它们就称为一个逆序。
一个排列中逆序的总数就称为这个排列的逆序数。逆序数为偶数的排列称为偶排列；逆序数为奇数的排列称为奇排列。
如2431中，21，43，41，31是逆序，逆序数是4，为偶排列。
思路：每次把末尾的数掉到序列前面时，减少的逆序对数为n-1-a[i] ，
增加的逆序对数为a[i] ，这样就可在所有的序列中找出含有逆序对最少的了！
#include <iostream>using namespace std;#define N 5005int main(){int a[N];int n,sum,i,j,min;while(scanf("%d",&n)!=EOF){sum=0;for(i=1;i<=n;i++)scanf("%d",&a[i]);for(i=1;i<n;i++){for(j=i+1;j<=n;j++){if(a[i]>a[j])sum++;}}min=sum;for(i=n;i>=1;i--){sum-=n-1-a[i];sum+=a[i];if(min>sum)min=sum;}printf("%d/n",min);}return 0;}