POJ 1410 Intersection（判断线段交和点在矩形内）

Intersection

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 14773 Accepted: 3872

Description

You are to write a program that has to decide whether a given line segment intersects a given rectangle. 

An example: 
line: start point: (4,9) 
end point: (11,2) 
rectangle: left-top: (1,5) 
right-bottom: (7,1) 

 
Figure 1: Line segment does not intersect rectangle 

The line is said to intersect the rectangle if the line and the rectangle have at least one point in common. The rectangle consists of four straight lines and the area in between. Although all input values are integer numbers, valid intersection points do not have to lay on the integer grid. 

Input

The input consists of n test cases. The first line of the input file contains the number n. Each following line contains one test case of the format: 
xstart ystart xend yend xleft ytop xright ybottom 

where (xstart, ystart) is the start and (xend, yend) the end point of the line and (xleft, ytop) the top left and (xright, ybottom) the bottom right corner of the rectangle. The eight numbers are separated by a blank. The terms top left and bottom right do not imply any ordering of coordinates.

Output

For each test case in the input file, the output file should contain a line consisting either of the letter "T" if the line segment intersects the rectangle or the letter "F" if the line segment does not intersect the rectangle.

Sample Input

14 9 11 2 1 5 7 1

Sample Output

F

题意:给个实心矩形和直线，判断是否相交

思路:判断直线是否在矩形内或者跟边界相交

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<cstring>#include<string>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)memset(a,b,sizeof(a))#define eps 1e-10#define inf 1e8int n,m,cntl,cntp;double add(double a,double b){    if(abs(a+b)<eps*(abs(a)+abs(b))) return 0;    return a+b;}struct P{    double x,y;    P(){}    P(double x,double y): x(x),y(y){}    P operator + (P p)    {        return P(add(x,p.x),add(y,p.y));    }    P operator - (P p)    {        return P(add(x,-p.x),add(y,-p.y));    }    P operator *(double d)    {        return P(x*d,y*d);    }    double dot (P p)    {        return add(x*p.x,y*p.y);    }    double det(P p)    {        return add(x*p.y,-y*p.x);    }}p[100005],q[100005];bool on_seg(P p1,P p2,P q){    return (p1-q).det(p2-q)==0&&(p1-q).dot(p2-q)<=0;}P intersection(P p1,P p2,P q1,P q2){    return p1+(p2-p1)*((q2-q1).det(q1-p1)/(q2-q1).det(p2-p1));}bool xiangjiao(P p1,P p2,P p3,P p4){    if((p1-p2).det(p3-p4)==0)    {        if(on_seg(p1,p2,p3)||on_seg(p1,p2,p4)||on_seg(p3,p4,p1)||on_seg(p3,p4,p2))        {            return true;        }        return false;    }    P r=intersection(p1,p2,p3,p4);    if(on_seg(p1,p2,r)&&on_seg(p3,p4,r))        return true;    return false;}int main(){    cin>>n;    while(n--)    {        double xs,xe,ys,ye,xl,yt,xr,yb;        double x1,x2,y1,y2;        scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&xs,&ys,&xe,&ye,&x1,&y1,&x2,&y2);        xl=min(x1,x2);        xr=max(x1,x2);        yb=min(y1,y2);        yt=max(y1,y2);        if(xiangjiao(P(xs,ys),P(xe,ye),P(xl,yt),P(xr,yt))||xiangjiao(P(xs,ys),P(xe,ye),P(xl,yb),P(xr,yb))||xiangjiao(P(xs,ys),P(xe,ye),P(xl,yb),P(xl,yt))||xiangjiao(P(xs,ys),P(xe,ye),P(xr,yb),P(xr,yt)))        {            cout<<"T"<<endl;        }        else if(max(xs, xe) < xr && max(ys, ye) < yt && min(xs, xe) > xl && min(ys, ye) > yb)        {            cout<<"T"<<endl;        }        else cout<<"F"<<endl;    }    return 0;}