HDU4720

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思路：

1.求三角形外切圆

2.求出圆点与半径，再判断给出的点是否在圆内

知识点：

1.已知三点求外接圆圆心： rx=(x1+x2+x3)/3,ry=(y1+y2+y3)/3

/*Code By Aquariuslt*//*HDU 4720 Naive and Silly Muggles*/#include<iostream>#include<stdio.h>#include<string>#include<algorithm>#include<string.h>#include<set>#include<vector>#include<queue>#include<stack>#include<limits.h>#include<math.h>#define FOR(i,a,b) for(int i=(a);i<(b);i++)#define FORD(i,a,b) for(int i=(a);i<=(b);i++)#define REP(i,b) FOR(i,0,b)#define CLR(a) memset(a,0,sizeof(a))using namespace std;#define maxn 10001int main(){    int t;    scanf("%d",&t);    REP(ti,t){        double x1,x2,x3,y1,y2,y3;        double lx,ly,rx,ry;        scanf("%lf %lf",&x1,&y1);        scanf("%lf %lf",&x2,&y2);        scanf("%lf %lf",&x3,&y3);        scanf("%lf %lf",&lx,&ly);        rx=(x1+x2+x3)/3.0;        ry=(y1+y2+y3)/3.0;        double r=sqrt((rx-x1)*(rx-x1)+(ry-y1)*(ry-y1));        double leng=sqrt((rx-lx)*(rx-lx)+(ry-ly)*(ry-ly));        if(r>=leng)printf("Case #%d: Danger\n",ti+1);        else printf("Case #%d: Safe\n",ti+1);    }    return 0;}