HDU-1394-Minimum Inversion Number （单点更新+区间求和）

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. 

InputThe 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. 
OutputFor 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：给一个0-n-1的排列，这个排列中的逆序数为数对 (ai, aj) 满足 i < j and ai > aj的个数。依次把第一个数放到排列的末尾会得到另外n-1个排列，求这n个排列中的最小的逆序数。

思路：首先要求第一个排列的逆序数。假如第一个逆序数为s0,当把a0从首位移到末位时，新得到的s1应该是在s0的基础上加上比a0大的数的个数，减去比a0小的数的个数。由于这一串数是一个0~n-1的排列，所以比a0大的数的个数为 (n - 1) - (a0 + 1) + 1 = n - a0 - 1，比a0小的数的个数为 (a0-1)-0+1=a0，所以s1 = s0 + n - a0 - a0 - 1.逆序数就是求对于每个数aj，在坐标比aj小的前提下，求比它大的数的个数。建树的时候把每个节点都初始化为0，每当插入一个数，就在这个数对应的叶子节点上加1，同时更新包含这个点的线段所对应的非叶子节点（加1），一层层更新上去，区间求和即可。

#include <cstdio>
#include <algorithm>
using namespace std;
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define mid (l+r)>>1
#define abbreviations int l,int r,int rt
const int maxn = 5005;
int sum[maxn<<2];
int a[maxn];
void PushUP(int rt)
{
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void build(abbreviations)
{
    sum[rt] = 0;
    if(l == r) return;
    int m = mid;
    build(lson);
    build(rson);
}
int query(int L,int R, abbreviations)
{
    if(L <= l && r <= R)
        return sum[rt];
    int m = mid;
    int ret = 0;
    if(L <= m) ret += query(L,R,lson);
    if(R>m) ret += query(L,R,rson);
    return ret;
}
void update(int p, abbreviations)
{
    if(l == r)
    {
        sum[rt]++;
        return;
    }
    int m = mid;
    if(p <= m) update(p,lson);
    else update(p,rson);
    PushUP(rt);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        build(0, n-1, 1);
        int sum = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += query(a[i], n-1, 0, n-1, 1);
            update(a[i], 0, n-1, 1);
        }
        int minm = sum;
        for(int i = 0; i < n; i++)
        {
            sum += n - a[i] - a[i] - 1;
            minm = min(minm, sum);
        }
        printf("%d\n", minm);
    }
    return 0;
}