[51nod]1298 圆与三角形

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https://www.51nod.com/onlineJudge/questionCode.html#problemId=1298&noticeId=54853

判断圆与三角形是否相交，转化为判断点到线段的最小距离。

1A

要整理模板

#include<cstring>#include<cstdio>#include<cmath>#include<iostream>#include<algorithm>using namespace std;const double eps=1e-8;struct node{    double x,y;    node(){}    node(int x_,int y_)    {        x=x_;        y=y_;    }    node operator -(const node &b)    {        return node(x-b.x,y-b.y);    }    double operator *(const node &b)    {        return x*b.x+y*b.y;    }    double operator ^(const node &b)    {        return x*b.y-b.x*y;    }} p,a[3];double r;int sgn(double x){    if(fabs(x)<eps) return 0;    if(x>0) return 1;    else return -1;}double dist(const node &a,const node &b){    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}double point_to_seg(node p,node a,node b){    double y=(p-a)*(b-a);    if(sgn(y)<=0) return dist(a,p);    if(sgn(y-dist(a,b))>=0) return dist(b,p);    double s=(p-a)^(b-a);    return s*s/dist(a,b);}bool ok(){    for(int i=0;i<3;i++)    {        int x=sgn(dist(a[i],p)-r*r);        int y=sgn(dist(a[(i+1)%3],p)-r*r);        if(x*y==0 || x*y==-1) return true;        if(x<0 && y<0) continue;        if(sgn(point_to_seg(p,a[i],a[(i+1)%3])-r*r)<=0) return true;    }    return false;}int main(){    int sk;    scanf("%d",&sk);    while(sk--)    {        scanf("%lf%lf%lf",&p.x,&p.y,&r);        for(int i=0;i<3;i++)        {            scanf("%lf%lf",&a[i].x,&a[i].y);        }        puts(ok()?"Yes":"No");    }}

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