Uva11343-isolated segments (two segments intersect)

Test instructions: give you some line segments to find the number of segments that don't intersect other segments

Formula: P1*p2= (X1*X2,Y1*Y2) (inner product), p1xp2= (x1*y2,x2*y1) (outer product)

Determine if q is on the line p1-p2 above, according to (P1-Q) x (p2-q) = "Q" is the linear p1-p2.

Use the inner product (p1-q) * (P2-Q) <0 to determine if q is between p1-p2.

Intersection of P1-P2,Q1-Q2:

(x, Y) =p1+ (P2-P1) * ((Q2-Q1) x (Q1-P1)/((Q2-Q1) x (P2-P1)));

Reasoning: The P1-P2 line is written in point P1+t (P2-P1), then point and Q1-q2 line, and finally calculate the above.

Finally, verify that the intersection is not on two segments.

This problem is mainly precision, look at the absolute value of <1000, in fact, when the calculation will appear more than int, so need to use a long long, if it is used double, pay attention to precision, because all are integers, EPS should be set a little bigger, Too small will be due to double precision and WA. Double Judge A==0, use ABS (a) <eps.

#include <iostream> #include <cstdio> #include <cmath> #include <vector> #include <list>#    include<cstring>using namespace std; #define EPS 1e-3#define N 1005int ans[n];struct point{Double A, B;    Point () {}-point (double a,double b): A (a), B (b) {}-operator + (point P) {return-point (P.A+A,P.B+B);    } Point operator – (point P) {return point (A-P.A,B-P.B);    } Point operator * (double p) {return point (a*p,b*p);    } double dot (point P)//inner product {return (P.A*A+P.B*B);    } double det (point P)//outer product {return (A*P.B-B*P.A);    }};struct seg{int index; Point P1,p2;}; Point p1,p2;vector<seg>v;//Determines if q is p1-p2 on the line segment bool On_str (p1,point p2,point q) {return (ABS (P1-Q). Det (p2-q)) & Lt eps&& (p1-q). dot (p2-q) <eps);} Ask for a two-line intersection point intersection (Point P1,point p2,point q1,point Q2) {return p1+ (P2-P1) * ((Q2-Q1). Det (Q1-P1)/(Q2-Q1). Det ( P2-P1));} void solve (int a) {for (int i=0;i<v.siZe (); i++) {point q1=v[i].p1;         Point Q2=v[i].p2; if (ABS (P1-P2) det (q1-q2)) <eps)//Determine if the outer product is 0 {if (on_str (p1,p2,q1) | | On_str (P1,P2,Q2) | | On_str (Q1,Q2,P1) | |                  On_str (Q1,Q2,P2))//Determine if there is a coincidence {ans[a]=1;              Ans[v[i].index]=1;             }} else {point r=intersection (P1,P2,Q1,Q2);                 if (On_str (p1,p2,r) &&on_str (q1,q2,r)) {ans[a]=1;             Ans[v[i].index]=1;    }}}}int Main () {int t;    scanf ("%d", &t);            while (t--) {int n;            V.clear ();            memset (ans,0,sizeof (ans));        scanf ("%d", &n);                    for (int i=1;i<=n;i++) {scanf ("%lf%lf%lf%lf", &p1.a,&p1.b,&p2.a,&p2.b);                Solve (i);        V.push_back (SEG{I,P1,P2});        } int t=0;      for (int i=1;i<=n;i++) {if (ans[i]==0)          t++;    } printf ("%d\n", t); } return 0;}

Uva11343-isolated segments (two segments intersect)