最小覆蓋圓的模板；

  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 //計算兩點距離
 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 //計算三角形面積
 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 //求三角形的外接圓，返回圓心和半徑(存在結構體"圓"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 //根據S = a * b * c / R / 4;求半徑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 //確定最小包圍圓
 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 //遞歸實現增量算法
 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 }

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轉載于:https://www.cnblogs.com/yours1103/p/3317640.html

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