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!