HDU 1394 Minimum Inversion Number

题目链接：https://vjudge.net/problem/HDU-1394

题目大意：

　　给定一个由 0~n-1 组成的长度为 n 序列，如果将最前面一个元素放到最后面去，就形成了一个新序列，新序列进行同样的操作可以再形成新序列，一共能形成 n 的新序列，每个序列都有一个逆序数，求其中最小的逆序数。

分析：

　　求出 a₁ ~ a_n 的逆序数后，其余序列的逆序数可在O(1)时间内推出。

代码如下：

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
 
#define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
#define Rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}
 
typedef long long LL;
typedef unsigned long long uLL;
typedef pair< double, double > PDD;
typedef pair< int, int > PII;
typedef set< int > SI;
typedef vector< int > VI;
typedef map< int, int > MII;
typedef vector< LL > VL;
typedef vector< VL > VVL;
const double EPS = 1e-10;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 5e3 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

int n, ans; 
int a[maxN];

#define lson l , mid , rt << 1 
#define rson mid + 1 , r , rt << 1 | 1

struct SegmentTree{
    int st[maxN << 2];
    
    inline void pushUp(int rt) {
        st[rt] = st[rt << 1] + st[rt << 1 | 1];
    }
    
    inline void pushDown(int rt) { }
    
    inline void build(int l, int r, int rt) {
        ms0(st);
    }
    
    // 单点替换，a[x] = y 
    inline void update(int x, LL y, int l, int r, int rt) {
        if(l >= r) {
            st[rt] = y;
            return;
        }
        int mid = (l + r) >> 1;
        if(x <= mid) update(x, y, lson);
        else update(x, y, rson);
        pushUp(rt);
    }
    
    // 区间查询 
    inline LL querySum(int L, int R, int l, int r, int rt) {
        if(L <= l && r <= R) return st[rt];
        LL ret = 0;
        int mid = (l + r) >> 1;
        
        if(L <= mid) ret += querySum(L, R, lson);
        if(R > mid) ret += querySum(L, R, rson);
        return ret;
    }
};
SegmentTree segTr;

int main(){
    INIT();
    while(cin >> n) {
        For(i, 1, n) {
            cin >> a[i];
            ++a[i];
        }
        ans = 0;
        segTr.build(1, n, 1);
        
        For(i, 1, n) {
            ans += segTr.querySum(a[i], n, 1, n, 1); // 计算大于a[i]且在a[i]之前出现过的数的个数 
            segTr.update(a[i], 1, 1, n, 1); // 标记为1，表示已经出现过 
        }
        
        int tmp = ans;
        For(i, 1, n - 1) {
            tmp = tmp + n - 2 * a[i] + 1;
            ans = min(ans, tmp);
        }
        cout << ans << endl;
    }
    return 0;
}