线段树——求逆序对hdu1394

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

题意：求一个排列的逆序对数，然后移动这个排列得到的新排列最小的逆序对

模板题

线段树区间内数字出现的次数和，当加入一个a[ i ]时查询他右边的数字出现的次数；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<set>
using namespace std;
const int maxn=5005;
int n;
int b[maxn<<2];
int a[maxn];
void build(int l,int r,int rt)
{
    b[rt]=0;
    if(l==r) return ;
    int mid=(l+r)>>1;
    build(l,mid,rt*2);
    build(mid+1,r,rt*2+1);
}
void add(int l,int r,int rt,int v)
{
    b[rt]+=1;
    if(l==r)
    {
        return ;
    }
    int mid=(l+r)>>1;
    if(v<=mid)    add(l,mid,rt*2,v);
    else    add(mid+1,r,rt*2+1,v);
}
int sum(int l,int r,int rt,int ll,int rr)
{
    if(ll<=l&&r<=rr) return b[rt];
    int mid=(l+r)>>1;
    int ans=0;
    if(ll<=mid)     ans+=sum(l,mid,rt<<1,ll,rr);
    if(rr>mid)    ans+=sum(mid+1,r,rt<<1|1,ll,rr);
    return ans;
}
int main()
{
    int v;
    while(~scanf("%d",&n))
    {
        build(1,n,1);
        int ans=0;
        for(int i=1;i<=n;++i)
            {
                scanf("%d",&a[i]);
                ++a[i];
            }
        for(int i=1;i<=n;i++)
        {
            ans+=sum(1,n,1,a[i],n);//1到a[i]右边界的元素个数即是所求
            add(1,n,1,a[i]);//a[i]右边界

        }
        int mmin=ans;
        for(int i=1;i<=n;++i)
        {
            ans-=a[i]-1;
            ans+=n-a[i];
            
            mmin=min(ans,mmin);
        }
        cout<<mmin<<endl;
    }
    return 0;
}