poj 1410 Intersection

题目描述良心大大地坏，大家一定要注意最后输入矩形的时候不一定是左上右下的输入，所以要做判断

根据题目意思，一条线段和矩形相交就是分成两部分：

1）矩形的四条边之一和这条线段相交

2）这条线段有一个点在矩形内

于是可以AC了

CODE：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>

using namespace std;
#define FOR(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define DOR(i,a,b) for(int (i)=(a);(i)>=(b);(i)--)
#define oo 1e6
#define eps 1e-8
#define nMax 1000
#define pb push_back
#define bug puts("OOOOh.....");
#define zero(x) (((x)>0?(x):-(x))<eps)

int dcmp(double x){
if(fabs(x)<eps) return 0;
return x>0?1:-1;
}
struct point {
double x,y;
point(double x=0,double y=0): x(x),y(y) {}
void make(double _x,double _y) {x=_x;y=_y;}
void read(){ scanf("%lf%lf",&x,&y); }
double len(){ return sqrt(x*x+y*y); }
friend point operator -(point const& u,point const& v) {
return point(u.x-v.x,u.y-v.y);
}
friend point operator +(point const& u,point const& v) {
return point(u.x+v.x,u.y+v.y);
}
friend double operator *(point const& u,point const& v) {
return u.x*v.y-u.y*v.x;
}
friend double operator ^(point const& u,point const& v) {
return u.x*v.x+u.y*v.y;
}
friend point operator *(double const& k,point const& v) {
return point(k*v.x,k*v.y);
}
friend point operator /(point const& u,double const& k){
return point(u.x/k,u.y/k);
}
friend bool operator < (point const& u,point const& v) {
if(dcmp(u.x-v.x) == 0) return dcmp(u.y-v.y)<0;
return dcmp(u.x-v.x)<0;
}
friend bool operator == (point const& u,point const& v) {
return (dcmp(u.x-v.x) == 0) && dcmp(u.y-v.y)==0;
}
friend int dots_online(point,point,point);
};
int dots_online(point a,point b,point c){ return dcmp((a-c)*(b-c))==0; }
typedef struct line{
point a,b;
line() {}
line(point a,point b): a(a),b(b) {}
void make(point _a,point _b) {a=_a;b=_b;}
void read() { a.read(),b.read(); }
friend int intersection(line,line);
} segment;
int dot_in_line(point p,line l){
return dcmp((l.a-p)*(p-l.b))==0 && dcmp((l.a-p)^(p-l.b))>=0;
}
int sameside(point a,point b,line l){
return dcmp((l.a-a)*(a-l.b)) * dcmp((l.a-b)*(b-l.b)) > 0;
}
int intersection(line u,line v){
if(dots_online(u.a,u.b,v.a) && dots_online(u.a,u.b,v.b))
return dot_in_line(u.a,v) || dot_in_line(u.b,v) || dot_in_line(v.a,u) || dot_in_line(v.a,u);
else
return !sameside(u.a,u.b,v) && !sameside(v.a,v.b,u);
}

line l,s[4];
point p1,p2,p3,p4;
int n;
double x1,x2,yy1,y2;

int inside(point a,point p1,point p2){
return a.x>=p1.x&&a.x<=p2.x&&a.y>=p2.y&&a.y<=p1.y;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("input.txt","r",stdin);
#endif
scanf("%d",&n);
while(n--){
l.read();cin>>x1>>yy1>>x2>>y2;
if(x1>x2) swap(x1,x2);
if(yy1>y2) swap(yy1,y2);
p1.make(x1,yy1);
p2.make(x1,y2);
p3.make(x2,y2);
p4.make(x2,yy1);
s[0].make(p1,p2),s[1].make(p2,p3),s[2].make(p3,p4),s[3].make(p4,p1);
int ok=1;
for(int i=0;i<4;i++) if(intersection(l,s[i]) || inside(l.a,p2,p4)) {ok=0;break;}
printf("%s\n",ok?"F":"T");
}
return 0;
}