HDOJ 1394 Minimum Inversion Number calculates the Minimum Number of reverse orders of cyclic strings (violent & amp; line segment tree)

HDOJ 1394 Minimum Inversion Number calculates the Minimum Number of reverse orders of cyclic strings (violent & line segment tree)
Minimum Inversion NumberTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission (s): 14879 Accepted Submission (s): 9082


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

 

 

 

Ac code:

Brute force solution: 425 ms

 

#include
 
  #include
  
   #include
   
    #include
    
     #include#define INF 0x7fffffff#define MAXN 10010#define max(a,b) a>b?a:b#define min(a,b) a>b?b:ausing namespace std;int num[MAXN];int main(){int i,j,n;while(scanf(%d,&n)!=EOF){int M;int cnt=0;for(i=0;i
     
      cnt)M=cnt;}printf(%d,M);}return 0;}
     
    
   
  
 

I can see that many people use line segment trees. I also want to write about them. tomorrow!