intersection--poj1410 (determining the relationship between a segment and a rectangle)

http://poj.org/problem?id=1410

The main topic: give you a line and a rectangle diagonal two points if the intersection of the output ' T ' Don't want to cross is ' F '

Attention:

1, given the rectangle may not be left upper and lower right so to judge first

2, the segment is in the inner output of the rectangle T

3, if the intersection is a vertex of a rectangle, it is disjoint.

Situation a little more just start thinking too little WA's I'm distressed

#include <stdio.h>#include<string.h>#include<stdlib.h>#include<ctype.h>#include<math.h>#include<queue>#include<algorithm>#include<iostream>using namespacestd;#defineN 20Const DoubleESP = 1e-8;#defineINF 0x3f3f3f3f#defineMemset (A, B) memset (A,b,sizeof (a))structpoint{Doublex, y; Point (Doublex=0,Doubley=0): X (x), Y (y) {} pointoperator- (ConstPoint &temp)Const    {        returnPoint (x-temp.x,y-temp.y); }    int operator* (ConstPoint &temp)Const    {        DoubleT= (X*TEMP.Y)-(y*temp.x); if(t>ESP)return 1; if(Fabs (T) <ESP)return 0; Else            return-1; }} p[n],j; Point Line (Point U1,point u2,point v1,point v2)///finding the intersection template{point ret=U1; DoubleT= ((u1.x-v1.x) * (V1.Y-V2.Y)-(U1.Y-V1.Y) * (v1.x-v2.x))/((u1.x-u2.x) * (V1.Y-V2.Y)-(U1.Y-U2.Y) * (v1.x-v2.x)); Ret.x+ = (u2.x-u1.x) *T; Ret.y+ = (U2.Y-U1.Y) *T; returnret;}intMain () {intN; scanf ("%d",&N);  while(n--) {j.x=j.y=INF; Memset (P,0); DoubleX1,x2,y1,y2; Doublea,b,c,d; scanf ("%lf%lf%lf%lf",&x1,&y1,&x2,&y2); scanf ("%lf%lf%lf%lf",&a,&b,&c,&d); if(a>c) Swap (A,C), swap (B,D); intu,d; U=Max (b,d); D=min (b,d); if(X1>=a && x1<=c && x2>=a && x2<=c && y1>=d && y1<=u && y 2>=d && y2<=u) {printf ("t\n"); Continue; }        ///determines whether a line segment intersects an edge of a rectanglep[1]=Point (X1,Y1); p[2]=Point (X2,y2); p[3]=Point (A, b); p[4]=Point (C,B); p[5]=Point (C,B); p[6]=Point (C,D); p[7]=Point (A, b); p[8]=Point (A,D); p[9]=Point (A,D); p[Ten]=Point (C,D); intK,kk; intflag=0;  for(intI=2; i<=5; i++) {k=abs ((p[1]-p[i*2-1]) * (p[i*2]-p[i*2-1]) + (p[2]-p[i*2-1]) * (p[i*2]-p[i*2-1])); KK=abs ((p[i*2-1]-p[1]) * (p[2]-p[1]) + (p[i*2]-p[1]) * (p[2]-p[1])); if(k!=2&& kk!=2)///if the intersection            {                if((p[1]-p[i*2-1]) * (p[i*2]-p[i*2-1])==0&& (p[2]-p[i*2-1]) * (p[i*2]-p[i*2-1])==0)///co-line                {                    if(i==2|| i==5)                    {                        if((p[1].x>a && p[1].X&LT;C) | | (p[2].x>a && p[2].X&LT;C) | | (p[1].x<=a && p[2].X&GT;=C) | | (p[2].x<=a && p[1].x>=c)) {flag=1;  Break; }                    }                    Else                    {                        if((p[1].y>d && p[1].y<u) | | (p[2].y>d && p[2].y<u) | | (p[1].y<=d && p[2].y>=u) | | (p[2].y<=d && p[1].y>=u)) {flag=1;  Break; } }} J=line (p[1],p[2],p[i*2],p[i*2-1]); if(i==2)                if(J.x>a && j.x<c && j.y==b)///if the intersection is between A and C (not including a C){flag=1;  Break; }                if(i==3)                {                    if(J.x==c && j.y>d && j.y<u) {flag=1;  Break; }                }                if(i==4)                {                    if(J.x==a && j.y>d && j.y<u) {flag=1;  Break; }                }                if(i==5)                {                    if(J.x>a && j.x<c && j.y==d) {flag=1;  Break; }                }            }        }        if(flag==0) printf ("f\n"); Elseprintf ("t\n"); }    return 0;}

intersection--poj1410 (determining the relationship between a segment and a rectangle)