51Nod1298 圆与三角形

题目链接：https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1298
题意：中文题
解析：首先要分情况讨论，三个点都在园内，那么就是不相交，三个点在圆外，那么问题就变成了圆与线段是否相交的问题了，先求点到直线距离，如果点到直线距离是大于半径的，那么这条直线肯定不相交，否则需要判断，圆心在这条直线上投影的位置，如果在线段外面，则说明圆与线段不相交，如果三条线段都与圆不相交，那么剩下的就是三角形与圆相交了
由于怕精度问题，全部都让他们

#include 
using namespace std;
typedef long long ll;
struct point
{
    ll x,y;
    point() {}
    point(ll _x,ll _y):x(_x),y(_y) {}
    point operator - (const point &b)const
    {
        return point(x-b.x,y-b.y);
    }
};
ll d_mul(point p1,point p2)
{
    return p1.x*p2.x+p1.y*p2.y;
}
ll dis(point p1,point p2)
{
    return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
}
ll judge(point p1,point p2,point p0,ll r)
{
    ll a,b,c;  //ax+by+c=0
    if(p1.x==p2.x)
        a = 1,b = 0,c = -p1.x;
    else if(p1.y==p2.y)
        a = 0,b = 1,c = -p1.y;
    else
    {
        a = p2.y-p1.y;
        b = p1.x-p2.x;
        c = p2.x*p1.y-p2.y*p1.x;
    }
    ll t1 = a*p0.x+b*p0.y+c;
    t1 *= t1;
    ll t2 = (a*a+b*b)*r*r;
    if(t1>t2)
        return false;
    t1 = d_mul(p0-p1,p2-p1);
    t2 = d_mul(p0-p2,p1-p2);
    if(t1>0 && t2>0)
        return true;
    else
        return false;
}
bool slove(point a,point b,point c,point p0,ll r)
{
    ll d1 = dis(a,p0);
    ll d2 = dis(b,p0);
    ll d3 = dis(c,p0);
    r *= r;
    if(d1return false;
    if(d1>r && d2>r && d3>r)
        return judge(a,b,p0,r) || judge(a,c,p0,r) || judge(b,c,p0,r);
    return true;
}
int main(void)
{
    int t;
    cin>>t;
    while(t--)
    {
        point p0,a,b,c;
        int r;
        cin>>p0.x>>p0.y>>r;
        cin>>a.x>>a.y;
        cin>>b.x>>b.y;
        cin>>c.x>>c.y;
        if(slove(a,b,c,p0,r))
            puts("Yes");
        else
            puts("No");
    }
    return 0;
}