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

101 3 6 9 0 8 5 7 4 2

 

本题较水，我用暴力过了，就不想打线段树！

AC代码（暴力实现）

#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);  //交换俩个不相邻的数a，b（a在前，b在后），逆序数+=两者间大于a的个数-两者间大于b的个数！！！  if(sum<ans)             ans=sum;             }          printf("%d\n",ans);     } return 0;}