看刘汝佳的训练指南过的第一道计算几何题。开始的时候看到这题就觉得这是水题，于是瞬间就把代码打出来了。不过在debug的时候，搞着搞着发现如果像我一开始的时候那样做，好多特殊情况不能正常处理，加了好多特判都还是能找出bug。最后在搞出下面那么多数据，觉得这样特判简直就是一个无底洞，根本就不能正确的得到答案。。。- - 最后在wa到忍不住的情况下，只好求助于佳哥的题解了。。
看了一下他的代码，瞬间觉得豁然开朗，原来之前那么的多的特判都可以通过这样的方法来消除。
简单说一下他的做法。首先，我们可以知道顶点数V+边数E-面数F=2这个基本定理。然后，这个图形的顶点数是可以在两两判相交的时候得到的，我用一个set来保存这些个数。边数是解这题的一个关键，边数至少有n-1条，这是显而易见的，然而某些边是被分割成多少段呢？这时候就需要将我们找到的点放进线段，看看有没有将一条线段分割，有的话就将总的边数加一。因为将点放进set里了，所以所有点都是唯一的。也就是不会有同一个位置的点对同一条线段分割多于一次。最后套进欧拉定理，就可以正确的把这题解决了。
代码如下：
View Code
1 #include <cstdio> 2 #include <cstring> 3 #include <cmath> 4 #include <set> 5 #include <vector> 6 #include <iostream> 7 #include <algorithm> 8 9 using namespace std; 10 11 #define REP(i, n) for (int i = 0; i < (n); i++) 12 13 struct Point { 14 double x, y; 15 Point() {} 16 Point(double x, double y) : x(x), y(y) {} 17 } ; 18 template<class T> T sqr(T x) { return x * x;} 19 20 // basic calculations 21 typedef Point Vec; 22 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);} 23 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);} 24 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);} 25 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);} 26 27 const double eps = 1e-8; 28 int sgn(double x) { return fabs(x) < eps ? 0 : (x < 0 ? -1 : 1);} 29 bool operator < (Point a, Point b) { return a.x < b.x || (a.x == b.x && a.y < b.y);} 30 bool operator == (Point a, Point b) { return sgn(a.x - b.x) == 0 && sgn(a.y - b.y) == 0;} 31 32 double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;} 33 double vecLen(Vec x) { return sqrt(sqr(x.x) + sqr(x.y));} 34 double angle(Vec a, Vec b) { return acos(dotDet(a, b) / vecLen(a) / vecLen(b));} 35 double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;} 36 double triArea(Point a, Point b, Point c) { return fabs(crossDet(b - a, c - a));} 37 Vec rotate(Vec x, double rad) { return Vec(x.x * cos(rad) - x.y * sin(rad), x.x * sin(rad) + x.y * cos(rad));} 38 Vec normal(Vec x) { 39 double len = vecLen(x); 40 return Vec(- x.y / len, x.x / len); 41 } 42 43 struct Line { 44 Point s, t; 45 Line() {} 46 Line(Point s, Point t) : s(s), t(t) {} 47 } ; 48 typedef Line Seg; 49 50 bool onSeg(Point x, Point a, Point b) { return sgn(crossDet(a - x, b - x)) == 0 && sgn(dotDet(a - x, b - x)) < 0;} 51 bool onSeg(Point x, Seg s) { return onSeg(x, s.s, s.t);} 52 // 0 : not intersect 53 // 1 : proper intersect 54 // 2 : improper intersect 55 int segIntersect(Point a, Point c, Point b, Point d) { 56 Vec v1 = b - a, v2 = c - b, v3 = d - c, v4 = a - d; 57 int a_bc = sgn(crossDet(v1, v2)); 58 int b_cd = sgn(crossDet(v2, v3)); 59 int c_da = sgn(crossDet(v3, v4)); 60 int d_ab = sgn(crossDet(v4, v1)); 61 if (a_bc * c_da > 0 && b_cd * d_ab > 0) return 1; 62 if (onSeg(b, a, c) && c_da) return 2; 63 if (onSeg(c, b, d) && d_ab) return 2; 64 if (onSeg(d, c, a) && a_bc) return 2; 65 if (onSeg(a, d, b) && b_cd) return 2; 66 return 0; 67 } 68 int segIntersect(Seg a, Seg b) { return segIntersect(a.s, a.t, b.s, b.t);} 69 70 // point of the intersection of 2 lines 71 Point lineIntersect(Point P, Vec v, Point Q, Vec w) { 72 Vec u = P - Q; 73 double t = crossDet(w, u) / crossDet(v, w); 74 return P + v * t; 75 } 76 Point lineIntersect(Line a, Line b) { return lineIntersect(a.s, a.t - a.s, b.s, b.t - b.s);} 77 78 // directed distance 79 double pt2Line(Point x, Point a, Point b) { 80 Vec v1 = b - a, v2 = x - a; 81 return crossDet(v1, v2) / vecLen(v1); 82 } 83 double pt2Line(Point x, Line L) { return pt2Line(x, L.s, L.t);} 84 85 double pt2Seg(Point x, Point a, Point b) { 86 if (a == b) return vecLen(x - a); 87 Vec v1 = b - a, v2 = x - a, v3 = x - b; 88 if (sgn(dotDet(v1, v2)) < 0) return vecLen(v2); 89 if (sgn(dotDet(v1, v3)) > 0) return vecLen(v3); 90 return fabs(crossDet(v1, v2)) / vecLen(v1); 91 } 92 double pt2Seg(Point x, Seg s) { return pt2Seg(x, s.s, s.t);} 93 94 struct Poly { 95 vector<Point> pt; 96 Poly() {} 97 Poly(vector<Point> pt) : pt(pt) {} 98 double area() { 99 double ret = 0.0; 100 int sz = pt.size(); 101 for (int i = 1; i < sz; i++) { 102 ret += crossDet(pt[i], pt[i - 1]); 103 } 104 return fabs(ret / 2.0); 105 } 106 } ; 107 108 /****************** template above *******************/ 109 110 Poly poly; 111 112 int main() { 113 // freopen("in", "r", stdin); 114 int cas = 1, n; 115 while (cin >> n && n) { 116 set<Point> pt; 117 pt.clear(); 118 poly.pt.clear(); 119 int ln = n - 1; 120 double x, y; 121 for (int i = 0; i < n; i++) { 122 cin >> x >> y; 123 poly.pt.push_back(Point(x, y)); 124 pt.insert(Point(x, y)); 125 for (int j = 1; j < i; j++) { 126 int st = segIntersect(Line(poly.pt[i - 1], poly.pt[i]), Line(poly.pt[j - 1], poly.pt[j])); 127 if (st == 1) { 128 pt.insert(lineIntersect(Line(poly.pt[i - 1], poly.pt[i]), Line(poly.pt[j - 1], poly.pt[j]))); 129 } 130 } 131 } 132 for (set<Point>::iterator si = pt.begin(); si != pt.end(); si++) { 133 for (int i = 1; i < n; i++) { 134 if (onSeg(*si, poly.pt[i - 1], poly.pt[i])) ln++; 135 } 136 } 137 cout << "Case " << cas++ << ": There are " << ln - pt.size() + 2 << " pieces." << endl; 138 } 139 return 0; 140 }
数据如下：
View Code
1 5 2 0 0 3 0 1 4 1 1 5 1 0 6 0 0 7 8 7 9 1 1 10 1 5 11 2 1 12 2 5 13 5 1 14 3 5 15 1 1 16 17 11 18 0 0 19 4 0 20 8 0 21 8 1 22 7 0 23 6 1 24 4 0 25 3 1 26 2 0 27 0 1 28 0 0 29 30 7 31 0 0 32 1 0 33 2 0 34 2 1 35 1 0 36 0 1 37 0 0 38 39 12 40 0 0 41 1 -1 42 2 0 43 1 1 44 0 0 45 2 0 46 2 1 47 0 1 48 0 0 49 -1 -1 50 0 -1 51 0 0 52 53 11 54 0 0 55 4 0 56 8 0 57 8 1 58 7 0 59 6 1 60 4 0 61 3 1 62 2 0 63 0 1 64 0 0 65 66 5 67 0 0 68 10 0 69 5 5 70 5 -5 71 0 0 72 73 7 74 0 0 75 0 -1 76 1 0 77 -1 0 78 1 -2 79 1 1 80 0 0 81 82 11 83 0 0 84 1 0 85 1 1 86 -1 -1 87 0 -1 88 0 1 89 -1 1 90 1 -1 91 0 -2 92 -2 0 93 0 0 94 95 14 96 0 0 97 1 0 98 1 1 99 -1 -1 100 0 -1 101 0 1 102 -1 1 103 1 -1 104 0 -2 105 -2 0 106 0 0 107 100 101 108 99 101 109 0 0 110 111 16 112 0 0 113 1 0 114 1 1 115 -1 -1 116 0 -1 117 0 1 118 -1 1 119 1 -1 120 0 -2 121 -2 0 122 0 0 123 100 101 124 99 101 125 -99 -101 126 -100 -101 127 0 0 128 129 0
——written by Lyon