HDU1394-Minimum Inversion Number

Minimum Inversion Number

                                                                            Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
                                                                                                     Total Submission(s): 18396    Accepted Submission(s): 11168

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

 

题意：给定一个长度为n的数组，那么每次将第一个元素放在最后一个能形成n个长度为n的数组，求出所有情况中逆序对最少的个数。

解题思路：设当前第一个元素为x，那么比它小的有x个元素，比它大的有n-x-1个元素，将它放到数组最后对逆序对的变化是n-2*x-1个。那么我们只要枚举这个分开的位置m就行了。

树状数组：

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;const int MAXN=5050;int f[MAXN],a[MAXN];;int n;int lowbit(int x){    return x&(-x);}void add(int x,int val){    while(x<=n)    {        f[x]+=val;        x+=lowbit(x);    }}int getsum(int x){    int s=0;    while(x>0)    {        s+=f[x];        x-=lowbit(x);    }    return s;}int main(){    while(~scanf("%d",&n))    {        int ans=0;        memset(f,0,sizeof(f));        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            a[i]++;            ans+=getsum(n)-getsum(a[i]);            add(a[i],1);        }        int mi=ans;        for(int i=1;i<=n;i++)        {            ans+=n-a[i]-(a[i]-1);            if(ans<mi) mi=ans;        }        printf("%d\n",mi);    }    return 0;}

暴力：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int main(){    int a[5009],sum[5009];    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d",&a[i]);        memset(sum,0,sizeof sum);        for(int i=0;i<n-1;i++)        {            for(int j=i+1;j<n;j++)                if(a[j]<a[i]) sum[i]++;        }        int ans=0;        for(int i=0;i<n;i++)            ans+=sum[i];        int t=ans;        for(int i=0;i<n-1;i++)        {            t=t-sum[i]+n-1-sum[i];            for(int j=i+1;j<n;j++)                if(a[i]<a[j]) sum[j]++;            if(t<ans) ans=t;        }        printf("%d\n",ans);    }    return 0;}