vjudge上题目链接：Determine the Shape
第二道独自 A 出的计算几何水题，题意就是给你四个点，让你判断它是哪种四边形：正方形、矩形、菱形、平行四边形、梯形 or 普通四边形。
按照各个四边形的特征层层判断就行，只是这题四个点的相对位置不确定（点与点之间是相邻还是相对并不确定），所以需要枚举各种情况，幸好 4 个点一共只有 3 种不同的情况：abcd、abdc、acbd 画个图就能一目了然，接下来就是编码了，无奈因为一些细节问题我还是调试了还一会 o(╯□╰)o
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<algorithm> 5 using namespace std; 6 7 struct Vector { 8 double x,y; 9 Vector(double x = 0, double y = 0): x(x), y(y) {} 10 void readint() { 11 int x,y; 12 scanf("%d %d",&x,&y); 13 this->x = x; 14 this->y = y; 15 } 16 Vector operator + (const Vector &b) const { 17 return Vector(x + b.x, y + b.y); 18 } 19 Vector operator - (const Vector &b) const { 20 return Vector(x - b.x, y - b.y); 21 } 22 Vector operator * (double p) const { 23 return Vector(x * p, y * p); 24 } 25 Vector operator / (double p) const { 26 return Vector(x / p, y / p); 27 } 28 }; 29 30 typedef Vector point; 31 typedef const point& cpoint; 32 typedef const Vector& cvector; 33 34 Vector operator * (double p, const Vector &a) { 35 return a * p; 36 } 37 38 double dot(cvector a, cvector b) { 39 return a.x * b.x + a.y * b.y; 40 } 41 42 double length(cvector a) { 43 return sqrt(dot(a,a)); 44 } 45 46 double cross(cvector a, cvector b) { 47 return a.x * b.y - a.y *b.x; 48 } 49 50 const double eps = 1e-10; 51 int dcmp(double x) { 52 return fabs(x) < eps ? 0: (x < 0 ? -1 : 1); 53 } 54 55 bool operator == (cpoint a, cpoint b) { 56 return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0; 57 } 58 59 const char s[8][30] = {"Ordinary Quadrilateral", "Trapezium", "Parallelogram", 60 "Rhombus", "Rectangle", "Square" }; 61 62 int judge(cpoint a, cpoint b, cpoint c, cpoint d) { 63 if(dcmp(cross(a - b, c - d)) == 0) { 64 if(dcmp(cross(b - c, a - d)) == 0) { 65 if(dcmp(length(b - a) - length(d - a)) == 0) { 66 if(dcmp(dot(b - a, d - a)) == 0) return 5; 67 else return 3; 68 } 69 else if(dcmp(dot(b - a, d - a)) == 0) return 4; 70 else return 2; 71 } 72 else return 1; 73 } 74 else if(dcmp(cross(b - c, a - d)) == 0) return 1; 75 else return 0; 76 } 77 78 int main() { 79 int t,Case = 0; 80 point a,b,c,d; 81 scanf("%d",&t); 82 while(t--) { 83 a.readint(); 84 b.readint(); 85 c.readint(); 86 d.readint(); 87 printf("Case %d: ",++Case); 88 int ans = 0; 89 ans = max(ans, judge(a,b,c,d)); 90 ans = max(ans, judge(a,b,d,c)); 91 ans = max(ans, judge(a,c,b,d)); 92 printf("%s ",s[ans]); 93 } 94 return 0; 95 }
计算几何，还是很有趣的。。。