http://blog.csdn.net/f_zyj/article/details/52066901
实在不好意思。看完了大神的博客，还是自己写着费劲哇,不好意思写个注释就说是原创。。当个模板先记下来吧。看两天晚上了。
说下学长给讲得这篇博客有些具体点的数学思路：
#include <iostream> #include <cstdio> using namespace std; typedef long long ll; typedef struct // 点结构 { ll x, y; } Point; Point A, B, C, O; // 三角形三点与圆心 ll r; // 半径<span style="white-space:pre"> </span> ll distan(Point a,Point b) //点到点的距离。 { return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y); } int solve(Point *p_1, Point *p_2) { ll a, b, c, dist_1, dist_2, angle_1, angle_2; //构造ax + by +c=0的形式 a = p_1->y - p_2->y; b = p_2->x - p_1->x; c = p_1->x * p_2->y - p_1->y * p_2->x; dist_1 = a * O.x + b * O.y + c; dist_1 *= dist_1; //这一步和下一步其实就是点到直线距离 与 半径的比较，不过为了防止/法的精度损失将下面的 a^2+b^2乘过去了 dist_2 = (a * a + b * b) * r * r; if (dist_1 > dist_2)<span style="white-space:pre"> </span> //如果R比距离要小，那么一定就不相交了 { return 0; }<span style="white-space:pre"> </span> //反之，如果不小，就不一定相交不相交，因为这个是线段与圆是否相交的问题，要用向量点乘积的夹角去考虑！ angle_1 = (O.x - p_1->x) * (p_2->x - p_1->x) + (O.y - p_1->y) * (p_2->y - p_1->y); angle_2 = (O.x - p_2->x) * (p_1->x - p_2->x) + (O.y - p_2->y) * (p_1->y - p_2->y); if (angle_1 > 0 && angle_2 > 0) { return 1; } return 0; } int intersect() { ll distance_a=distan(O,A); ll distance_b=distan(O,B); ll distance_c=distan(O,C); ll dist=r*r; if(distance_a<dist&&distance_b<dist&&distance_c<dist)//在圆内必定不相交 return 0; if(distance_a>dist&&distance_b>dist&&distance_c>dist) //三点都在圆外 { if(solve(&A,&B)||solve(&A,&C)||solve(&B,&C))//里面注释 return 1; return 0; } return 1;//其他情况必定相交，一点在内，一点在外 } int main() { int t; scanf("%d", &t); while (t--) { scanf("%lld%lld%lld%lld%lld%lld%lld%lld%lld", &O.x, &O.y, &r, &A.x, &A.y, &B.x, &B.y, &C.x, &C.y); printf("%s\n", intersect() ? "Yes" : "No"); } return 0; }