HDOJ 1394 Minimum Inversion Number 求循環串的最小逆序數（暴力&&線段樹）

Minimum Inversion Number

Time 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

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

 

 

 

ac代碼：

暴力解法：425ms

 

#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:a
using namespace std;
int num[MAXN];
int main()
{
	int i,j,n;
	while(scanf(%d,&n)!=EOF)
	{
		int M;
		int cnt=0;
		for(i=0;icnt)
			M=cnt;
		}
		printf(%d
,M);
	}
	return 0;
}

看到很多人都用線段樹來寫，我也想寫一下，明天吧！