HDU 1221 Rectangle and Circle

Rectangle and Circle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1933    Accepted Submission(s): 451

Problem Description

Given a rectangle and a circle in the coordinate system(two edges of the rectangle are parallel with the X-axis, and the other two are parallel with the Y-axis), you have to tell if their borders intersect.
Note: we call them intersect even if they are just tangent. The circle is located by its centre and radius, and the rectangle is located by one of its diagonal.

 

Input

The first line of input is a positive integer P which indicates the number of test cases. Then P test cases follow. Each test cases consists of seven real numbers, they are X,Y,R,X1,Y1,X2,Y2. That means the centre of a circle is (X,Y) and the radius of the circle is R, and one of the rectangle's diagonal is (X1,Y1)-(X2,Y2).

 

Output

For each test case, if the rectangle and the circle intersects, just output "YES" in a single line, or you should output "NO" in a single line.

 

Sample Input

2

1 1 1 1 2 4 3

1 1 1 1 3 4 4.5

 

Sample Output

YES

NO

 

Author

weigang Lee

 

Source

杭州电子科技大学第三届程序设计大赛

 

Recommend

Ignatius.L

 

 

 

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

using namespace std;

#define eps 1e-8

struct point{
    double x,y;
};

double dis(point p1,point p2){
    return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}

double xmult(point p1,point p2,point p0){
    return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);
}

double distoline(point p,point l1,point l2){
    return fabs(xmult(p,l1,l2)/dis(l1,l2));
}

int isIntersect(point c,double r,point l1,point l2){
    double t1=dis(c,l1)-r, t2=dis(c,l2)-r;
    point t=c;
    if(t1<eps || t2<eps)
        return t1>-eps || t2>-eps;
    t.x+=l1.y-l2.y;
    t.y+=l2.x-l1.x;
    return xmult(l1,c,t)*xmult(l2,c,t)<eps && distoline(c,l1,l2)-r<eps;
}

point p[4],cir;
double X,Y,R,X1,Y1,X2,Y2;

int main(){

    //freopen("input.txt","r",stdin);

    int t;
    cin>>t;
    while(t--){
        cin>>X>>Y>>R>>X1>>Y1>>X2>>Y2;
        p[0].x=X1;
        p[0].y=Y1;

        p[1].x=X1;
        p[1].y=Y2;

        p[2].x=X2;
        p[2].y=Y2;

        p[3].x=X2;
        p[3].y=Y1;

        cir.x=X;
        cir.y=Y;

        int flag=0;
        for(int i=0;i<4;i++)
            if(isIntersect(cir,R,p[i%4],p[(i+1)%4])){
                flag=1;
                break;
            }
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}