hdoj 1394 Minimum Inversion Number(树状数组求逆序对）

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

101 3 6 9 0 8 5 7 4 2

 

Sample Output

16

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

sum记录初始的逆序对数

之后对sum逆序循环操作  循环到i表示处理将第i个数从数列末尾调到开头的情况

总共有n个数，对于数a[i]在数列末尾时前面比他大的个数为n-a[i]-1，比他小的个数为a[i] 

当把a[i]调到开头时原先比他大的（n-a[i]-1)个所构成的逆序对变成正序 故sum-n+a[i]+1

原先比a[i]小的a[i]个正序对变成逆序对 故（sum-n+a[i]+1)+a[i]

循环一遍 取最小的sum 就是答案

#include<cstdio>#include<cstring>using namespace std;#define lowbit(x) x&(-x)#define N 5000+5#define INF 1000000000int cnt[N];int add(int n){    while(n<=N)    {        cnt[n]++;        n+=lowbit(n);    }    return 0;}int query(int n){    int sum=0;    while(n>0)    {        sum+=cnt[n];        n-=lowbit(n);    }    return sum;}int main(){    int a[N],n,sum,i,min;    while(scanf("%d",&n)!=EOF)    {        min=INF;        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        sum=0;        memset(cnt,0,sizeof(cnt));        for(i=1;i<=n;i++)        {            add(a[i]+1);            sum+=i-query(a[i]+1);        }        for(i=n;i>1;i--)        {            sum=sum-n+1+2*a[i];            if(sum<min)                min=sum;        }        printf("%d\n",min);    }    return 0;}