HDU - 1394 Minimum Inversion Number【单点增减，区间求和】-优快云博客

本文介绍了一种算法，用于解决给定序列通过不同数量的循环移位后得到的序列中，逆序对数目最少的问题。该算法使用了区间树进行优化处理，能够有效地计算出所有可能序列的逆序数，并找出最小值。

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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

#include<iostream>
#include<cstdio>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1|1
using namespace std;
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(int l, int r, int rt)
{
    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]++;
        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)
{
    if(L <= l && R >= r){
        return sum[rt];
    }
    int m = (l + r) >> 1;
    int ret = 0;
    if(L <= m) ret += query(L, R, lson);
    if(R > m) ret += query(L, R, rson);
    return ret;
}
int main()
{
    int n;
    while(~scanf("%d", &n)){
            build(0, n - 1, 1);
            int cnt = 0;
        for (int i = 0; i < n; i++){
                scanf("%d", &a[i]);
                cnt += query(a[i], n - 1, 0, n - 1, 1);
           update(a[i], 1, n - 1, 1);
        }
        int ans = cnt;
        for (int i = 0; i < n ; i++){
            cnt += n - a[i] - a[i] - 1;
            ans = min(cnt, ans);
         }
         cout << ans << endl;
    }

}