Hdu1542 line segment tree + scanning line + discretization

I just want to say whether the question is actually unacceptable or not. I suggest you refer to the one I wrote earlier.

# Include <stdio. h>

# Include <string. h>
# Include <algorithm>
# Include <iostream>
Using namespace STD;
# Define LL (x) (x <1)
# Define RR (x) (x <1) | 1)


Int N;
Struct Node
{
Double Y;
Int x1, x2;
Int flash;
} A [31000];
Struct Node
{
Int II;
Double X;
} Change [1, 35000];
Int CMP1 (node A, Node B)
{
Return A. x <B. X;
}
Int cmp2 (node A, Node B)
{
Return A. Y <B. Y;
}
Int num [80000];
Double map [31000];
Int Update (int l, int R, int left, int right, int K, int mark)
{
Int mid = (L + r)/2;
If (L = left & Right = r)
{
If (Num [Mark]! =-1)
{
Num [Mark] + = K;
Return 0;
}
Update (L, mid, L, mid, K, LL (Mark ));
Update (MID, R, mid, R, K, RR (Mark ));
If (Num [LL (Mark)] = num [RR (Mark)]) num [Mark] = num [LL (Mark)];
Else num [Mark] =-1;
}
Else
{
If (Num [Mark]> = 0)
{
Num [LL (Mark)] = num [RR (Mark)] = num [Mark];
}
If (right <= mid)
{
Update (L, mid, left, right, K, LL (Mark ));
}
Else if (left> = mid)
{
Update (MID, R, left, right, K, RR (Mark ));
}
Else
{
Update (L, mid, left, mid, K, LL (Mark ));
Update (MID, R, mid, right, K, RR (Mark ));
}
If (Num [LL (Mark)] = num [RR (Mark)]) num [Mark] = num [LL (Mark)];
Else num [Mark] =-1;
}
Return 0;
}
Double find (int l, int R, int mark)
{
Double long = 0;
If (Num [Mark]> = 1)
{
Long + = map [R]-map [l];
Return long;
}
If (Num [Mark] = 0) return 0;
Int mid = (L + r)/2;
Long + = find (L, mid, LL (Mark) + find (MID, R, RR (Mark ));
Return long;
}
Int main ()
{
Int I, j, D = 1;
Double X1, Y1, X2, Y2;
While (~ Scanf ("% d", & N), n)
{
N * = 2;
// J = 0;
Memset (A, 0, sizeof ());
Memset (MAP, 0, sizeof (MAP ));
Memset (change, 0, sizeof (Change ));
For (I = 1; I <= N; I + = 2)
{
Scanf ("% lf", & X1, & Y1, & X2, & Y2 );
A [I]. Y = Y1;
A [I]. Flash = 1;
A [I + 1]. Y = Y2;
A [I + 1]. Flash =-1;
Change [I]. x = x1;
Change [I + 1]. II = change [I]. II = I;
Change [I + 1]. x = x2;
}
Int leap [10000];
Memset (LEAP, 0, sizeof (LEAP ));
Sort (Change + 1, change + 1 + N, CMP1 );
Int J = 0;
Double K =-1;
For (I = 1; I <= N; I ++)
{
If (change [I]. X! = K)
{
K = change [I]. X;
++ J;
Map [J] = change [I]. X;
}
If (LEAP [change [I]. ii] = 0)
{
A [change [I]. II]. X1 = A [change [I]. II + 1]. X1 = J;
Leap [change [I]. ii] = 1;
}
Else a [change [I]. II]. X2 = A [change [I]. II + 1]. X2 = J;
}
Sort (a + 1, A + 1 + N, cmp2 );
Memset (Num, 0, sizeof (Num ));
Update (1, J, a [1]. X1, a [1]. X2, 1, 1 );
Double area = 0;
For (I = 2; I <= N; I ++)
{
Area + = find (1, J, 1) * (a [I]. Y-A [I-1]. y );
If (A [I]. Flash = 1)
Update (1, J, a [I]. X1, a [I]. X2, 1, 1 );
Else update (1, J, a [I]. X1, a [I]. X2,-1, 1 );
}
Printf ("Test Case # % d \ n", d ++ );
Printf ("total received area: %. 2lf \ n", area );
Printf ("\ n ");
}
Return 0;
}

Hdu1542 line segment tree + scanning line + discretization