51nod 1298 圆与三角形

来源：互联网发布：手机脚本怎么编程编辑：程序博客网时间：2024/05/16 10:59
点这里
计算几何
题不难。。
好想长一个脑子啊
感谢大佬相助
#include <iostream>#include <stdio.h>#include <math.h>#include <stdlib.h>using namespace std;double x[4],y[4];double a,b,r,A,B,C;double  dis(double x1,double y1,double x2,double y2){    return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));}int pd(double x1,double y1,double x2,double y2){    double x,y,z;    x=dis(x1,y1,a,b);    y=dis(x2,y2,a,b);    z=dis(x1,y1,x2,y2);         if(y*y+z*z-x*x<=0)             return 0;        else if(-y*y+z*z+x*x<=0)             return 0;       else        return 1;}int main(){   //freopen("in1.txt","r",stdin);    int n;    while(cin>>n)    {        while(n--)        {            int f1=1;            scanf("%lf%lf%lf",&a,&b,&r);            int t=0,z=0;            for(int i=0; i<3; i++)            {                scanf("%lf%lf",&x[i],&y[i]);                double tmp=(x[i]-a)*(x[i]-a)+(y[i]-b)*(y[i]-b);                if(tmp<r*r)                    t++;                if(tmp>r*r)                    z++;            }           // printf("%d %d\n",t,z);            if(t==3)                f1=0;            else if(z==3)            {         f1=0;                for(int i=0; i<2; i++)                {                    for(int j=i+1; j<3; j++)                    {                        //printf("%lf %lf %lf %lf\n",x[i],x[j],y[i],y[j]);                        double mm=0;                        if(x[i]==x[j])                        {                            if(fabs(a-x[i])<=r&&pd(x[i],y[i],x[j],y[j]))                            {                                f1=1;                            }                        }                        else if(y[i]==y[j])                        {                            if(fabs(b-y[i])<=r&&pd(x[i],y[i],x[j],y[j]))                            {                                f1=1;                            }                        }                        else                        {                            A=y[j]-y[i];                            B=x[i]-x[j];                            C=y[i]*(x[j]-x[i])-x[i]*(y[j]-y[i]);                            mm=fabs(A*a+B*b+C)/sqrt(A*A+B*B);                            //printf("***********************%lf %d\n",mm,f1);                            if(mm<=r&&pd(x[i],y[i],x[j],y[j]))                            {                                f1=1;                            }                        }                    }                }            }            else                f1=1;            // printf("f1=%d f2=%d\n",f1,f2);            if(f1)                puts("Yes");            else                puts("No");        }    }    return 0;}
阅读全文
0 0