1 #include<cstdio> 2 #include<cmath> 3 #include<queue> 4 #include<cstring> 5 #include<algorithm> 6 using namespace std; 7  8 const double eps = 1e-9; 9 const int N = 90001;10 11 struct point{//点12     double x, y;13     point(double _x = 0, double _y = 0): x(_x), y(_y){}14 };15 16 struct lineSegment{//线段17     point s, e;18     lineSegment(point s, point e): s(s), e(e){}19 };20 struct line{//直线21     double a, b, c;22 };23 bool operator < (point p1, point p2){24     return p1.x < p2.x || p1.x == p2.x && p1.y < p2.y;25 }26 bool operator == (point p1, point p2){27     return abs(p1.x - p2.x) < eps && abs(p1.y - p2.y) < eps;28 }29 bool onLine(lineSegment l, point p){//判断点是否在线段上30     return abs((l.e.x - l.s.x)*(p.y - l.s.y) - (p.x - l.s.x)*(l.e.y - l.s.y)) < eps //点在直线上31     && (p.x - l.s.x) * (p.x - l.e.x) < eps && (p.y - l.s.y) * (p.y - l.e.y) < eps;//点在线段内32 }33 line make_Line(point p1, point p2){//将线段延长为直线34     line l;35     l.a = (p2.y > p1.y) ? p2.y - p1.y : p1.y - p2.y;36     l.b = (p2.y > p1.y) ? p1.x - p2.x : p2.x - p1.x;37     l.c = (p2.y > p1.y) ? p1.y * p2.x - p1.x * p2.y : p1.x * p2.y - p1.y * p2.x;38     return l;39 }40 //判断直线是否相交，并求交点p41 bool line_Intersect(line l1, line l2, point &p){42     double d = l1.a * l2.b - l2.a * l1.b;43     if(abs(d) < eps) return false; //叉积为0,平行或重合44     p.x = (l2.c * l1.b - l1.c * l2.b) /d;45     p.y = (l2.a * l1.c - l1.a * l2.c) /d;46     return true;47 }48 //判断线段是否相交49 bool lineSegment_Intersect(lineSegment l1, lineSegment l2, point &p){50     line a, b;51     a = make_Line(l1.s, l1.e);//将线段延长为直线52     b = make_Line(l2.s, l2.e);53     if(line_Intersect(a, b, p))//如果直线相交54         //判断直线交点是否在线段上，是则线段相交55         return onLine(l1, p) && onLine(l2, p);56     else return false;57 }58 59 point p[N], intersection[N];60 int n, m;61 62 int main(){63     int nn, i, j, kase = 1;64     while(scanf("%d", &nn) && nn){65         n = m = 0;66         for(i = 0; i < nn; ++i)67             scanf("%lf %lf", &p[i].x, &p[i].y);68         for(i = 0; i < nn; ++i){69             for(j = 0; j < nn; ++j){70                 if(i == j) continue;71                 lineSegment l1(p[i], p[(i+1)%nn]), l2(p[j], p[(j+1)%nn]);72                 point v;73                 if(lineSegment_Intersect(l1, l2, v))74                     intersection[n++] = v;//记录交点75             }76         }77         sort(intersection , intersection + n);78         //移除重复点79         n = unique(intersection, intersection + n) - intersection;80         for(i = 0; i < n; ++i){81             for(j = 0; j < nn; ++j){82                 lineSegment l3(p[j], p[(j+1)%nn]);83                 //若有交点落在边上，则该边分裂成两条边84                 if(onLine(l3, intersection[i]) && !(l3.s == intersection[i]))85                     m++;86             }87         }88         printf("Case %d: There are %d pieces.\n", kase++, 2 + m - n);89     }90     return 0;91 }