HDU 1394 minimum inversion number (line segment tree)

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1394

The line segment tree calculates the reverse order number. First, the reverse order number (n logn) of the initial array is obtained, and then O (1) releases other Reverse orders. Because the input is 0 -- n-1, the build starts from 0.

Code:

# Include <cstdlib>
# Include <cctype>
# Include <cstring>
# Include <cstdio>
# Include <cmath>
# Include <algorithm>
# Include <vector>
# Include <string>
# Include <iostream>
# Include <sstream>
# Include <set>
# Include <queue>
# Include <stack>
# Include <fstream>
# Include <iomanip>
# Include <bitset>
# Include <list>
# Include <ctime>
Using namespace STD;

# Define set (ARR, what) memset (ARR, what, sizeof (ARR ))
# Define ff (I, A) for (I = 0; I <A; I ++)
# Define SD (a) scanf ("% d", &)
# Define SSD (a, B) scanf ("% d", & A, & B)
# Define SF (a) scanf ("% lf", &)
# Define SS (a) scanf ("% s",)
# Define SLD (a) scanf ("% LLD", &)
# Define PF (a) printf ("% d \ n",)
# Define PPF (a, B) printf ("% d \ n", a, B)
# Define SZ (ARR) (INT) A. Size ()
# Define swap (A, B) A = a xor B; B = a xor B; A = a xor B;
# Define read freopen ("in.txt", "r", stdin)
# Define write freopen ("out.txt", "W", stdout)
# Define max 1 <30
# Define ESP 1e-5
# Define lson L, M, RT <1 | 1
# Define rson m + 1, R, (RT <1) + 2
Template <class T> inline t sqr (t a) {return a * ;}
Template <class T> inline void Amin (T & A, t B) {if (a =-1 | A> B) A = B ;}
Template <class T> inline void amax (T & A, t B) {if (a <B) A = B ;}
Template <class T> inline T min (t a, t B) {return A> B? B: ;}
Template <class T> inline T max (t a, t B) {return A> B? A: B ;}
Const int maxn = 5010;
Int sum [maxn <2];
Int data [maxn];
Void build (int l, int R, int RT ){
Sum [RT] = 0;
If (L = r) return;
Int M = (L + r)> 1;
Build (lson );
Build (rson );
}
Void Update (int x, int L, int R, int RT ){
Sum [RT] ++;
If (L = r) return;
Int M = (L + r)> 1;
If (x <= m) Update (x, lson );
Else Update (x, rson );
}
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, I, j, INI, ans;
While (~ SD (N )){
Build (0, n-1, 0 );
INI = 0;
Ff (I, n ){
SD (data [I]);
INI + = query (data [I], n, 0, n-1, 0 );
Update (data [I], 0, n-1, 0 );
}
Ans = ini;
Ff (I, n ){
INI + = N-data [I]-data [I]-1;
Ans = min (INI, ANS );
}
PF (ANS );
}
Return 0;
}