hdu 4720
最小覆盖圆的模板;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<stdio.h> 2 #include<string.h> 3 #include<math.h> 4 struct Point 5 { 6 double x; 7 double y; 8 } pt[1005]; 9 struct Traingle 10 { 11 struct Point p[3]; 12 }; 13 struct Circle 14 { 15 struct Point center; 16 double r; 17 } ans; 18 //计算两点距离 19 double Dis(struct Point p, struct Point q) 20 { 21 double dx=p.x-q.x; 22 double dy=p.y-q.y; 23 return sqrt(dx*dx+dy*dy); 24 } 25 //计算三角形面积 26 double Area(struct Traingle ct) 27 { 28 return 29 fabs((ct.p[1].x-ct.p[0].x)*(ct.p[2].y-ct.p[0].y)-(ct.p[2].x-ct.p[0].x)*(ct.p[1].y-ct.p[0].y))/2.0; 30 } 31 //求三角形的外接圆,返回圆心和半径(存在结构体"圆"中) 32 struct Circle CircumCircle(struct Traingle t) 33 { 34 struct Circle tmp; 35 double a, b, c, c1, c2; 36 double xA, yA, xB, yB, xC, yC; 37 a = Dis(t.p[0], t.p[1]); 38 b = Dis(t.p[1], t.p[2]); 39 c = Dis(t.p[2], t.p[0]); 40 //根据S = a * b * c / R / 4;求半径R 41 tmp.r = (a*b*c)/(Area(t)*4.0); 42 xA = t.p[0].x; 43 yA = t.p[0].y; 44 xB = t.p[1].x; 45 yB = t.p[1].y; 46 xC = t.p[2].x; 47 yC = t.p[2].y; 48 c1 = (xA*xA+yA*yA - xB*xB-yB*yB) / 2; 49 c2 = (xA*xA+yA*yA - xC*xC-yC*yC) / 2; 50 tmp.center.x = (c1*(yA - yC)-c2*(yA - yB)) / ((xA - xB)*(yA - yC)-(xA - xC)*(yA - yB)); 51 tmp.center.y = (c1*(xA - xC)-c2*(xA - xB)) / ((yA - yB)*(xA - xC)-(yA - yC)*(xA - xB)); 52 return tmp; 53 } 54 //确定最小包围圆 55 struct Circle MinCircle(int num, struct Traingle ct) 56 { 57 struct Circle ret; 58 if (num==0) ret.r = 0.0; 59 else if (num==1) 60 { 61 ret.center = ct.p[0]; 62 ret.r = 0.0; 63 } 64 else if (num==2) 65 { 66 ret.center.x = (ct.p[0].x+ct.p[1].x)/2.0; 67 ret.center.y = (ct.p[0].y+ct.p[1].y)/2.0; 68 ret.r = Dis(ct.p[0], ct.p[1])/2.0; 69 } 70 else if(num==3) ret = CircumCircle(ct); 71 return ret; 72 } 73 //递归实现增量算法 74 void Dfs(int x, int num, struct Traingle ct) 75 { 76 int i, j; 77 struct Point tmp; 78 ans = MinCircle(num, ct); 79 if (num==3) return; 80 for (i=1; i<=x; i++) 81 if (Dis(pt[i], ans.center)>ans.r) 82 { 83 ct.p[num]=pt[i]; 84 Dfs(i-1, num+1, ct); 85 tmp=pt[i]; 86 for (j=i; j>=2; j--) 87 pt[j]=pt[j-1]; 88 pt[1]=tmp; 89 } 90 } 91 void Solve(int n) 92 { 93 struct Traingle ct; 94 Dfs(n, 0, ct); 95 } 96 int main (void) 97 { 98 int n, i,t; 99 int ca=1; 100 scanf("%d",&t); 101 while (t--) 102 { 103 for (i=1; i<=3; i++) 104 scanf("%lf %lf", &pt[i].x, &pt[i].y); 105 Solve(3); 106 double xx,yy; 107 scanf("%lf%lf",&xx,&yy); 108 printf("Case #%d: ",ca++); 109 if((xx-ans.center.x)*(xx-ans.center.x)+(yy-ans.center.y)*(yy-ans.center.y)<=(ans.r)*(ans.r)) 110 puts("Danger"); 111 else puts("Safe"); 112 } 113 return 0; 114 }