hdu 1394/哈理工oj 1250 Minimum Inversion Number【线段树过】

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15931    Accepted Submission(s): 9716

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
   
   
   
   

 

做这个题首先要能百度到这个东西：将序列中某一个数放到最后的逆序数改变量：n-2*a[i]-1;至于推导过程，呵呵哒就好~

思路：5000个数，多次查找当前数据前边有多少个数比他大，所以这里相当于多次询问过程，所以暴力过基本上是一定会超时的。

解法：线段树/树状数组求出当前序列的逆序数，然后根据百度到的这个结论解出最终解：
每一次输入一个数的时候、query一下前边有多少个数大于他，然后update它，搞定~

#include<stdio.h>
#include<string.h>
using namespace std;
#define lson l,m,rt*2
#define rson m+1,r,rt*2+1
int tree[21212121];
void pushup(int rt)
{
    tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void build( int l ,int r , int rt )
{
	if( l == r )
	{
		tree[rt]=0;
		return ;
	}
	else
	{
		int m = (l+r)>>1 ;
		build(lson) ;
		build(rson) ;
		pushup(rt) ;
	}
}
int Query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return tree[rt];
    }
    else
    {
        int m=(l+r)>>1;
        int ans=0;
        if(L<=m)
        {
            ans+=Query(L,R,lson);
        }
        if(m<R)
        {
            ans+=Query(L,R,rson);
        }
        return ans;
    }
}
void update(int p,int l,int r,int rt)
{
    if(l==r)
    {
        tree[rt]=1;
        return ;
    }
    else
    {
        int m=(l+r)/2;
        if(p<=m)update(p,lson);
        else update(p,rson);
        pushup(rt);
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        build(1,n,1);
        int s[100000];
        int sum=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&s[i]);
            sum+=Query(s[i]+1,n,1,n,1);
            update(s[i]+1,1,n,1);
        }
        //printf("%d\n",sum);
        int minn=sum;
        for(int i=0;i<n;i++)
        {
            sum=sum+n-2*s[i]-1;
            if(sum<minn)minn=sum;
        }
        printf("%d\n",minn);
    }
}