HDU 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9821    Accepted Submission(s): 6060

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
解题思路：树状数组求逆序数后递推求最小
#include <iostream>#include <cstdio>#include <cstring>#define lowbit(x) x&(-x)#define Maxn 5005using namespace std;int tree[Maxn];void add(int pos,int n){for(int i=pos;i<=n;i+=lowbit(i))tree[i]++;}int sum(int pos){int ans=0;for(int i=pos;i>0;i-=lowbit(i))ans+=tree[i];return ans;}int main(){int n,x[Maxn],ans;freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);while(~scanf("%d",&n)){ans=0;memset(tree,0,sizeof(tree));for(int i=0;i<n;i++){scanf("%d",&x[i]),x[i]++,add(x[i],n);ans+=x[i]-1-sum(x[i]-1);}int t=ans;for(int i=0;i<n;i++){ans=min(ans,t+n-2*x[i]+1);t=t+n-2*x[i]+1;}printf("%d\n",ans);}return 0;}