HDU 1394 - Minimum Inversion Number

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

 

大致题意：

　　一个由0..n-1组成的序列，每次可以把队首的元素移到队尾；
      求形成的n个序列中最小逆序对数目；

解题思路：

以下有两份代码：

　　代码一：

　　　　暴力法，复杂度 O(n^2);

　　　　当吧 a[0] 移到队尾后，会减少 a[0] 个逆序对 同时会增加  (n-1)-a[0] 个逆序对；

　　　　那么只要求出初始的逆序对数好了，剩下的可以推导得出；

　　代码二：

　　　　树状数组，复杂度0(nlogn);

　　　　无非就是用树状数组求逆序对，注意数列从0开始的，要处理过；

　　　　剩下的还是找规律；

 

#include <cstdio>
using namespace std;
int main(){
    int n,a[5000+5],t;
    while(~scanf("%d",&n)){
        for(int i=0;i<n;i++) scanf("%d",&a[i]);
        t=0;
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
                if(a[i]>a[j]) t++;
        int min=t;
        for(int i=0;i<n;i++){
            t=t-a[i]+(n-1)-a[i];
            if(t<min) min=t;
        }
        printf("%d\n",min);
    }
} 

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define N 5005
int c[N],a[N],n,m;
int modify(int x,int num){while(x<=n)c[x]+=num,x+=x&-x;}
int sum(int x){int s=0;while(x>0)s+=c[x],x-=x&-x;return s;}
int main(){
    while(~scanf("%d",&n)){
        memset(c,0,sizeof(c));
        m=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            a[i]+=1;//0->n-1 >>> 1->n
            modify(a[i],1);
            m+=i+1-sum(a[i]);//0->n-1 >>> 1->n
        } int ans=m;
        for(int i=0;i<n;i++){
            a[i]-=1;//变回来
            ans=min(ans,m=m-a[i]+n-1-a[i]);
        } printf("%d\n",ans);
    } return 0;
}