在别人的几何练习中找出来的简单的几何加二分的题目。题意是，给出隔槽的构造，统计出给出的点在哪一个隔槽中，最后将所有的统计结果输出来。
主要是不能用cin cout输入输出大量数据。这里用double也不会超时，不过用int会快很多。
代码如下：
1 #include <cstdio> 2 #include <cstring> 3 #include <vector> 4 #include <cmath> 5 #include <iostream> 6 #include <algorithm> 7 8 using namespace std; 9 10 struct Point { 11 double x, y; 12 Point() {} 13 Point(double x, double y) : x(x), y(y) {} 14 } ; 15 template<class T> T sqr(T x) { return x * x;} 16 inline double ptDis(Point a, Point b) { return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));} 17 18 typedef Point Vec; 19 Vec operator + (Vec a, Vec b) { return Vec(a.x + b.x, a.y + b.y);} 20 Vec operator - (Vec a, Vec b) { return Vec(a.x - b.x, a.y - b.y);} 21 Vec operator * (Vec a, double p) { return Vec(a.x * p, a.y * p);} 22 Vec operator / (Vec a, double p) { return Vec(a.x / p, a.y / p);} 23 24 const double EPS = 1e-10; 25 const double PI = acos(-1.0); 26 inline int sgn(double x) { return (x > EPS) - (x < -EPS);} 27 bool operator < (Point a, Point b) { return sgn(a.x - b.x) < 0 || sgn(a.x - b.x) == 0 && a.y < b.y;} 28 bool operator == (Point a, Point b) { return sgn(a.x - b.x) == 0 && sgn(a.y - b.y) == 0;} 29 30 inline double dotDet(Vec a, Vec b) { return a.x * b.x + a.y * b.y;} 31 inline double crossDet(Vec a, Vec b) { return a.x * b.y - a.y * b.x;} 32 inline double crossDet(Vec o, Vec a, Vec b) { return crossDet(a - o, b - o);} 33 inline bool onSeg(Point x, Point a, Point b) { return sgn(crossDet(x, a, b)) == 0 && sgn(dotDet(a - x, b - x)) < 0;} 34 35 int ptInPoly(Point p, Point *poly, int n) { 36 int wn = 0; 37 poly[n] = poly[0]; 38 for (int i = 0; i < n; i++) { 39 if (onSeg(p, poly[i], poly[i + 1])) return -1; 40 int k = sgn(crossDet(poly[i], poly[i + 1], p)); 41 int d1 = sgn(poly[i].y - p.y); 42 int d2 = sgn(poly[i + 1].y - p.y); 43 if (k > 0 && d1 <= 0 && d2 > 0) wn++; 44 if (k < 0 && d2 <= 0 && d1 > 0) wn--; 45 } 46 return wn != 0; 47 } 48 49 const int N = 5555; 50 double x[N][2], X1, Y1, X2, Y2; 51 int cnt[N]; 52 Point p[10]; 53 54 void insert(double X, double Y, int n) { 55 int h = 0, t = n, m, ok = -1; 56 Point cur = Point(X, Y); 57 while (h <= t) { 58 m = h + t >> 1; 59 p[0] = Point(X1, Y2); 60 p[1] = Point(X1, Y1); 61 p[2] = Point(x[m][1], Y1); 62 p[3] = Point(x[m][0], Y2); 63 bool front = ptInPoly(cur, p, 4); 64 if (front) ok = m, t = m - 1; 65 else h = m + 1; 66 } 67 if (ok == -1) { puts("shit~"); return ;} 68 cnt[ok]++; 69 } 70 71 int main() { 72 int n, m; 73 bool pnt = false; 74 while (cin >> n && n) { 75 cin >> m; 76 if (pnt) puts(""); 77 pnt = true; 78 cin >> X1 >> Y1 >> X2 >> Y2; 79 if (X1 > X2) swap(X1, X2); 80 if (Y1 > Y2) swap(Y1, Y2); 81 for (int i = 0; i < n; i++) { 82 for (int j = 0; j < 2; j++) { 83 scanf("%lf", &x[i][j]); 84 } 85 } 86 x[n][0] = x[n][1] = X2; 87 memset(cnt, 0, sizeof(cnt)); 88 double x, y; 89 for (int i = 0; i < m; i++) { 90 scanf("%lf%lf", &x, &y); 91 insert(x, y, n); 92 } 93 for (int i = 0; i <= n; i++) printf("%d: %d\n", i, cnt[i]); 94 } 95 return 0; 96 } 97 98
——written by Lyon