HDU 1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7907    Accepted Submission(s): 4851

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

 

Recommend

Ignatius.L

 

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数状数组维护。。。。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;#define nMax 5010int a[nMax],n;int c[nMax];int lowbit(int x) {return x&(-x);}void add(int x,int k) {while(x<=n) {c[x] += k;x += lowbit(x);}}int sum(int x) {int ret=0;while(x>0) {ret += c[x];x -= lowbit(x);}return ret;}int main() {//freopen("input.txt","r",stdin);while(~scanf("%d",&n)) {int ret=0;memset(c,0,sizeof(c));for(int i=0;i<n;i++) {scanf("%d",&a[i]);ret += sum(n) - sum(a[i]);add(a[i]+1,1);}int ans = ret;for(int i=0;i<n;i++) {int res = ret - a[i];res += n-a[i]-1;if(res < ans) ans=res;ret = res;}printf("%d\n",ans);}return 0;}