【线段树】单点更新3

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用线段树求逆序数

对于a1,a2,....an 的逆序数sum

我们求a2,a3,....an,a1的逆序数时，

假设比a1小的数的个数为t,则比a1大的数为n-t-1；

 那么sum=sum+n-2*t-1;

由于这题给出的数据是0~n-1的数，所以不需要离散化，比ai小的数的个数就是ai个

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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 
       
       

 

#include<cstdio>
using namespace std;

const int maxn=5005;

int a[maxn<<2];

void build(int l, int r, int rt)
{
    if(l==r)
    {
        a[rt]=0;
        return;
    }
    int mid=(l+r)>>1;
    build(l, mid, rt<<1);
    build(mid+1, r, rt<<1|1);
    a[rt]=0;
}

void PushUp(int rt)
{
    a[rt]=a[rt<<1]+a[rt<<1|1];
}

void updata(int l, int r, int x, int rt)
{
    if(l==r)
    {
        a[rt]++;
        return;
    }
    int m=(l+r)>>1;
    if(x<=m) updata(l, m, x, rt<<1);
    else updata(m+1, r, x, rt<<1|1);
    PushUp(rt);
}

int Query(int l, int r, int L, int R, int rt)
{
    if(L<=l && r<=R)
    {
        return a[rt];
    }
    int m=(l+r)>>1;
    int ans1=0, ans2=0;
    if(L<=m) ans1+=Query(l, m, L, R, rt<<1);
    if(R>m)  ans2+=Query(m+1, r, L, R, rt<<1|1);
    return ans1+ans2;
}

int main()
{
    int n, i, sum;
    int b[5005];
    while(scanf("%d", &n)!=EOF)
    {
        sum=0;
        build(0, n-1, 1);
        for(i=0; i<n; i++)
        {
            scanf("%d", &b[i]);
            sum+=Query(0, n-1, b[i], n-1, 1);
            updata(0, n-1, b[i], 1);
        }
        int min=sum;
        for(i=0; i<n; i++)
        {
            sum=sum+n-2*b[i]-1;
            if(sum<min)
            min = sum;
        }
        printf("%d\n", min);
    }



    return 0;
}