Poj 1514 Metal Cutting semi-flat cross plane cutting

Half plane intersection template O (N ^ 2) in http://www.cnblogs.com/wuyiqi/archive/2012/03/30/2426175.html

Question: Judge the sum of the shortest cut marks of the convex polygon required to be cut out on a square.

There are only eight sides in total, and the data size is small. The time limit is 10 s. The order of brute force enumeration and cutting is calculated.

This is my first contact with plane cutting. It can be called a virgin cutting.

View code

# Include <cmath>
# Include <cstdio>
# Include <algorithm>
 Using   Namespace STD;
 Const    Int Maxn = 100 ;
 Const   Double EPS = 1E- 8 ;
Inline Double Min ( Double A, Double B ){ Return A <B? A: B ;}
Inline Double SGN ( Double X ){ Return FABS (x) <EPS? 0 :( X> 0 ? 1 :- 1 );}
 Struct Point {
 Double X, Y;
Point ( Double Tx = 0 , Double Ty = 0 ) {X = TX; y = ty ;}
 Bool   Operator = ( Const Point & T) Const {
 Return SGN (x-t.x) =0 & SGN (y-t.y) = 0 ;
}
} P [maxn], set [maxn], St [maxn], TMP [maxn], PP [maxn];
 Struct SEG {point S, E ;};
Inline Double Dist (point a, point B ){ Return SQRT (A. x-b.x) * (A. x-b.x) + (A. y-b.y) * (A. y-b.y ));}
Inline Double Cross (point a, point B, point C ){ Return (B. x-a.x) * (C. y-a.y)-(B. y-a.y) * (C. x-a.x );}
Inline Bool Outside (SEG seg, point P ){ Return Cross (SEG. S, SEG. E, P)> EPS ;}
Inline Bool Inside (SEG seg, point P ){ Return Cross (SEG. S, SEG. E, P) <-EPS ;}
Point intersect (point P1, point P2, point P3, point P4, point & P ){
 Double A1, B1, C1, A2, B2, C2, D;
A1 = p1.y-p2.y; b1 = p2.x-p1.x; C1 = p1.x * p2.y-p2.x * p1.y;
A2 = p3.y-p4.y; b2 = p4.x-p3.x; C2 = p3.x * p4.y-p4.x * p3.y;
D = A1 * B2-A2 * B1;
 If (FABS (d) <EPS)Return   False ;
P. x = (-C1 * B2 + C2 * B1)/d;
P. Y = (-A1 * C2 + A2 * C1)/d;
 Return P;
}
 Double W, h;
 Int A [ 10 ], N, PN;
 Double Cut (SEG seg, point P []) {
 Int I, j, TOT = 0 ;
Point A, B;
A = B = point ( 0 , 0 );
 Bool S, E;
 For (I = 0 ; I <PN; I ++ ){
 If (! Outside (SEG, P [I]) PP [tot ++] = P [I];
 Else {
 If (I = 0 &&! Outside (SEG, P [pn- 1 ]) {B =;
PP [tot ++] = intersect (SEG. S, SEG. E, P [I], p [pn-1 ], );
}
 If (I! = 0 &&! Outside (SEG, P [I- 1 ]) {B =;
PP [tot ++] = intersect (SEG. S, SEG. E, P [I], p [I- 1 ], );
}
 If (! Outside (SEG, P [I + 1 ]) {
B =;
PP [tot ++] = intersect (SEG. S, SEG. E, P [I], p [I + 1 ], );
}
}
}
PP [tot] = PP [ 0 ]; Pn = tot;
Memcpy (St, PP, Sizeof (Pp ));
 Return Dist (A, B );
}
 Int Main (){
 Int I;
 While (Scanf ( "  % Lf  " , & W, & H )! = EOF ){
 Double Ans = 1e20;
St [ 4 ] = ST [ 0 ] = Point ( 0 , 0 );
St [ 1 ] = Point ( 0 , H );
St [ 2 ] = Point (W, H );
St [ 3 ] = Point (W, 0 );
Memcpy (TMP, St,Sizeof (St ));
Scanf ( "  % D  " , & N); seg ts [ 100 ];
 For (I = 0 ; I <n; I ++ ){
Scanf ( "  % Lf  " , & P [I]. X, & P [I]. y );
A [I] = I;
} P [N] = P [ 0 ];
For (I = 0 ; I <n; I ++) ts [I]. S = P [I], TS [I]. E = P [I + 1 ];
 Do {
 Double Tlen = 0 ;
Memcpy (St, TMP, Sizeof (TMP ));
Pn = 4 ;
 For (I = 0 ; I <n; I ++ ){
Tlen + = cut (TS [A [I], St );
}
Ans = min (ANS, tlen );
} While (Next_permutation (A, A + n ));
Printf ( "  Minimum total length = %. 3lf \ n  " , ANS );
}
 Return   0 ;
}