POJ 4048 Chinese repeating Crossbow segment specification intersection judgment

Test instructions

For n segments and initial points, the maximum number of lines emitted by the initial point can be crossed (even through the intersection).

Analysis:

Enumerating the end points of a segment, judging whether the segments intersect, requires a rigorous template.

Code:

POJ 4048//sep9 #include <iostream> using namespace std;
typedef long Long LL;
const LL maxn=2000;

const LL maxl=40028; struct P {ll x, y;}

A[maxn],b[maxn],p;
ll Tx,ty;

int n;
	ll det (P a,p b,p c) {ll x1=b.x-a.x;
	ll Y1=b.y-a.y;
	ll x2=c.x-a.x;
	ll Y2=c.y-a.y;
return x1*y2-x2*y1;
	} int Get_sign (ll x) {if (x==0) return 0;
Return x>0?1:-1; } ll Mymax (ll A,ll b) {return a>b?a:b;} ll mymin (ll A,ll b) {return a>b?b:a;} int between (P mid,p a,p b) {R Eturn Mid.x<=max (a.x,b.x) &&mid.y<=max (A.Y,B.Y) &&mid.x>=min (a.x,b.x) &&mid.y> 
=min (A.Y,B.Y);
 	} bool Crossed (P a,p b,p c,p d) {int d1,d2,d3,d4;
 	D1=get_sign (Det (a,b,c));
 	D2=get_sign (Det (a,b,d));
 	D3=get_sign (Det (c,d,a));	 
	D4=get_sign (Det (c,d,b));
	if ((D1^D2) ==-2&& (D3^D4) ==-2) return true;
	   if ((D1==0&&between (c,a,b)) | |
	   (D2==0&&between (d,a,b)) | |
	   (D3==0&&between (a,c,d)) | |
		(D4==0&&between (b,c,d)))
	return true; return FAlse;
	} int Process (P dir) {int cnt=0;
	dir.x=p.x+ (dir.x-p.x) *10000;
	dir.y=p.y+ (DIR.Y-P.Y) *10000;
	if (DIR.X==P.X&AMP;&AMP;DIR.Y==P.Y) return 0;
	for (int i=1;i<=n;++i) if (crossed (P,dir,a[i],b[i])) ++cnt;
return CNT;
	} int main () {int i,cases;
	scanf ("%d", &cases);
		while (cases--) {scanf ("%d", &n);
		for (I=1;i<=n;++i) scanf ("%lld%lld%lld%lld", &a[i].x,&a[i].y,&b[i].x,&b[i].y);			
		scanf ("%lld%lld", &p.x,&p.y);
		int ans=0;
			for (i=1;i<=n;++i) {Ans=max (ans,process (A[i]));
		Ans=max (Ans,process (b[i));
	} printf ("%d\n", ans);
} return 0;  }