Minimum Inversion Number

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.

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.

For each case, output the minimum inversion number on a single line.

101 3 6 9 0 8 5 7 4 2

16

CHEN, Gaoli

Ignatius.L   |   We have carefully selected several similar problems for you:            

暴力解法

#include
    
     #include
     
      #include
      
       #include
       
        using namespace std;int a[5500];int n;int main(){	while(~scanf("%d",&n))  //总共最多有n-1种逆数对 	{		for(int i=0;i
        
         a[j])				ans++;			}		}		int res=999999999;		if(ans
        
       
      
     
    

线段树解法

#include
    
     #include
     
      #include
      
       #include
       
         using namespace std; #define N 5555#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1int sum[N<<2];void pushup(int rt){	sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt)//初始化 {	sum[rt]=0;	if(l==r)	{		return;	}	int m=(l+r)>>1;	build(lson);	build(rson);	pushup(rt);}void update(int p,int l,int r,int rt){	if(l==r)	{		sum[rt]++;//由于存入了p这个点 sum[rt]的数量要加一  在递归更新节点的sum值 		return ;	}	int m=(l+r)>>1;	if(p<=m)update(p,lson);	else update(p,rson);	pushup(rt);}int query(int L,int R,int l,int r,int rt)//将L到R之间有多少点存在的数量 求出来 {	if(L<=l&&R>=r)	{		return sum[rt];	}	int m=(l+r)>>1;	int res=0;	if(L<=m)res+=query(L,R,lson);	if(R>m)res+=query(L,R,rson);	return res;}int x[N];int main(){	int n;	while(~scanf("%d",&n))	{		build(0,n-1,1);		int sum=0;		for(int i=0;i