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# HDU 1394 Minimum Inversion Number (树状数组求逆序

HDU 1394 Minimum Inversion Number (树状数组求逆序数)

Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13942 Accepted Submission(s): 8514
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
Author CHEN, Gaoli Source ZOJ Monthly, January 2003

#include
#include
#include
using namespace std;
int const MAX = 5005;
int c[MAX], a[MAX], n;

int lowbit(int x)
{
return x & (-x);
}

{
for(int i = x; i <= n; i += lowbit(i))
c[i] ++;
}

int Sum(int x)
{
int res = 0;
for(int i = x; i > 0; i -= lowbit(i))
res += c[i];
return res;
}

int main()
{
while(scanf(%d, &n) != EOF)
{
memset(c, 0, sizeof(c));
int ini = 0;
for(int i = 1; i <= n; i++)
{
scanf(%d, &a[i]);
a[i] ++;
ini += Sum(n) - Sum(a[i]);
}
int mi = ini;
for(int i = 1; i <= n; i++)
{
ini += (-(a[i] - 1) + n - (a[i] - 1) - 1);
mi = min(mi, ini);
}
printf(%d
, mi);
}
}

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