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：

  1 #pragma GCC optimize("Ofast")
  2 #include <bits/stdc++.h>
  3 using namespace std;
  4  
  5 #define INIT() std::ios::sync_with_stdio(false);std::cin.tie(0);
  6 #define Rep(i,n) for (int i = 0; i < (n); ++i)
  7 #define For(i,s,t) for (int i = (s); i <= (t); ++i)
  8 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)
  9 #define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
 10 #define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
 11 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
 12 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 13  
 14 #define pr(x) cout << #x << " = " << x << "  "
 15 #define prln(x) cout << #x << " = " << x << endl
 16  
 17 #define LOWBIT(x) ((x)&(-x))
 18  
 19 #define ALL(x) x.begin(),x.end()
 20 #define INS(x) inserter(x,x.begin())
 21  
 22 #define ms0(a) memset(a,0,sizeof(a))
 23 #define msI(a) memset(a,inf,sizeof(a))
 24 #define msM(a) memset(a,-1,sizeof(a))
 25 
 26 #define MP make_pair
 27 #define PB push_back
 28 #define ft first
 29 #define sd second
 30  
 31 template<typename T1, typename T2>
 32 istream &operator>>(istream &in, pair<T1, T2> &p) {
 33     in >> p.first >> p.second;
 34     return in;
 35 }
 36  
 37 template<typename T>
 38 istream &operator>>(istream &in, vector<T> &v) {
 39     for (auto &x: v)
 40         in >> x;
 41     return in;
 42 }
 43  
 44 template<typename T1, typename T2>
 45 ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
 46     out << "[" << p.first << ", " << p.second << "]" << "
";
 47     return out;
 48 }
 49  
 50 typedef long long LL;
 51 typedef unsigned long long uLL;
 52 typedef pair< double, double > PDD;
 53 typedef pair< int, int > PII;
 54 typedef set< int > SI;
 55 typedef vector< int > VI;
 56 typedef map< int, int > MII;
 57 typedef vector< LL > VL;
 58 typedef vector< VL > VVL;
 59 const double EPS = 1e-10;
 60 const int inf = 1e9 + 9;
 61 const LL mod = 1e9 + 7;
 62 const int maxN = 5e3 + 7;
 63 const LL ONE = 1;
 64 const LL evenBits = 0xaaaaaaaaaaaaaaaa;
 65 const LL oddBits = 0x5555555555555555;
 66 
 67 int n, ans; 
 68 int a[maxN];
 69 
 70 #define lson l , mid , rt << 1 
 71 #define rson mid + 1 , r , rt << 1 | 1
 72 
 73 struct SegmentTree{
 74     int st[maxN << 2];
 75     
 76     inline void pushUp(int rt) {
 77         st[rt] = st[rt << 1] + st[rt << 1 | 1];
 78     }
 79     
 80     inline void pushDown(int rt) { }
 81     
 82     inline void build(int l, int r, int rt) {
 83         ms0(st);
 84     }
 85     
 86     // 单点替换，a[x] = y 
 87     inline void update(int x, LL y, int l, int r, int rt) {
 88         if(l >= r) {
 89             st[rt] = y;
 90             return;
 91         }
 92         int mid = (l + r) >> 1;
 93         if(x <= mid) update(x, y, lson);
 94         else update(x, y, rson);
 95         pushUp(rt);
 96     }
 97     
 98     // 区间查询 
 99     inline LL querySum(int L, int R, int l, int r, int rt) {
100         if(L <= l && r <= R) return st[rt];
101         LL ret = 0;
102         int mid = (l + r) >> 1;
103         
104         if(L <= mid) ret += querySum(L, R, lson);
105         if(R > mid) ret += querySum(L, R, rson);
106         return ret;
107     }
108 };
109 SegmentTree segTr;
110 
111 int main(){
112     INIT();
113     while(cin >> n) {
114         For(i, 1, n) {
115             cin >> a[i];
116             ++a[i];
117         }
118         ans = 0;
119         segTr.build(1, n, 1);
120         
121         For(i, 1, n) {
122             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 
123             segTr.update(a[i], 1, 1, n, 1); // 标记为1，表示已经出现过 
124         }
125         
126         int tmp = ans;
127         For(i, 1, n - 1) {
128             tmp = tmp + n - 2 * a[i] + 1;
129             ans = min(ans, tmp);
130         }
131         cout << ans << endl;
132     }
133     return 0;
134 }

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