就是卡精度， 我把sqrt() 都去了， 换成平方与平方之间的比较， AC
思路：
1. 题目可以转换成判断线段与圆的关系。
那么你只要找到线段上离圆心最近的点，然后计算它与圆心的距离是否小于半径。
2. 要特判，线段是否包含于圆，是则线段依然不与圆交。
代码：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const double EP = 1e-8;
struct Point{
double x,y;
}p1 , p2 , p3, p4 ;
struct Cirlce{
double x,y,r;
}c;
double cal( double x, double y ){//点到圆心的距离的平方
return (x-c.x)*(x-c.x) + (y-c.y)*(y-c.y);
}
double sol_3x(double p , double Left, double Right ){
while( Left + EP < Right ){
double m1 = ( Left*2.0 + Right ) / 3.0;
double m2 = ( Left + Right*2.0 ) / 3.0;
if( cal( p ,m1 ) > cal( p ,m2 ))
Left = m1;
else Right = m2;
}
return Left;
}
double sol_3y(double p , double Left, double Right ){
while( Left + EP < Right ){
double m1 = ( Left*2.0 + Right ) / 3.0;
double m2 = ( Left + Right*2.0 ) / 3.0;
if( cal( m1 , p ) > cal( m2, p ))
Left = m1;
else Right = m2;
}
return Left;
}
bool isIntersect( Point p1, Point p2 ){
double most;
if( cal(p1.x,p1.y) < c.r*c.r && cal(p2.x,p2.y) < c.r*c.r ) return false;
if( p1.x == p2.x )
{
if( p1.y > p2.y ) swap( p1, p2 );
most = sol_3x( p1.x , p1.y , p2.y );
if( cal(p1.x,most ) <= c.r*c.r ) return true;
}
else {
if( p1.x > p2.x ) swap(p1,p2);
most = sol_3y( p1.y, p1.x, p2.x);
if( cal(most,p1.y) <= c.r*c.r ) return true;
}
return false;
}
void solve(){
p2.x = p3.x; p2.y = p1.y;
p4.x = p1.x; p4.y = p3.y;
if( ( isIntersect( p1, p2) || isIntersect( p2,p3) || isIntersect(p3,p4) || isIntersect(p4,p1) )) {puts("YES");return;}
puts("NO");
return ;
}
int main(){
int n; scanf("%d",&n);
while(n--){
scanf("%lf%lf%lf%lf%lf%lf%lf",&c.x,&c.y,&c.r,&p1.x,&p1.y,&p3.x,&p3.y ); solve();
}
return 0;
}