HDU1394 Minimum Inversion Number (树状数组)

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21638    Accepted Submission(s): 12934

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

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

思路：模拟一下就会发现，对于每个排列的第一个数而言，如果该数是排列中第x大的数，则在排列中存在x-1个数比该数小，存在n-x个数 比该数大，设当该数为排列的第一个元素时的逆序数为num，则当要把该数移动到排列末尾时，则当前的逆序数为 num + n - x - (x - 1),

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5010;int tree[maxn],a[maxn],n;int lowbit(int k){return k & -k;}void add(int num, int k){while(k <= n){tree[k] += num;k += lowbit(k);}}int query(int k){int sum = 0;while(k){sum += tree[k];k -= lowbit(k);}return sum;}int main(){while(~scanf("%d",&n)){for(int i = 1; i <= n; i ++) scanf("%d",&a[i]);memset(tree,0,sizeof(tree));int sum = 0;for(int i = 1; i <= n; i ++){a[i] ++;sum += query(n-a[i]+1);add(1,n-a[i]+1); }int ans = sum;for(int i = 1; i <= n; i ++){sum += n - 2 * a[i] + 1;ans = min(sum,ans);}printf("%d\n",ans);}return 0;}