POJ 3608 Bridge Across Islands(旋转卡壳)
转载:http://www.hankcs.com/program/algorithm/poj-3608-bridge-across-islands.html
跨岛大桥：在两个凸包小岛之间造桥，求最小距离？
1 #include <iostream> 2 #include <vector> 3 #include <algorithm> 4 #include <cmath> 5 using namespace std; 6 7 #define MAX_N 10000 + 16 8 #define INF 0x3F3F3F3F 9 #define EPS 1E-10 10 11 struct Point 12 { 13 double x, y; 14 Point() {} 15 Point(double x, double y) : x(x), y(y) {} 16 Point operator + (const Point& p){ return Point(x + p.x, y + p.y); } 17 Point operator - (const Point& p){ return Point(x - p.x, y - p.y); } 18 Point operator * (const double& d){ return Point(x * d, y * d); } 19 bool operator < (const Point& a) const 20 { 21 if (x != a.x) return x < a.x; 22 else return y < a.y; 23 } 24 double dot(const Point& p) { return x * p.x + y * p.y; } 25 double det(const Point& p) { return x * p.y - y * p.x; } 26 }; 27 28 Point P[MAX_N], Q[MAX_N]; 29 30 // 向量AB 与 AC 的叉积 如果叉积大于0，那么C在向量AB的逆时针方向，叉积小于0则在AB的顺时针方向。如果叉积等于0，则ABC共线。 31 inline double cross(Point A, Point B, Point C) 32 { 33 return (B - A).det(C - A); 34 } 35 36 // 向量AB 与 AC 的点积 如果点积的结果为0，那么这两个向量互相垂直 37 inline double multi(Point A, Point B, Point C) 38 { 39 return (B - A).dot(C - A); 40 } 41 42 // 两点距离 43 inline double dist(Point A, Point B) 44 { 45 return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y)); 46 } 47 48 // 逆时针排序 49 inline void anticlockwise_sort(Point* p, int N) 50 { 51 for (int i = 0; i < N - 2; ++i) 52 { 53 double tmp = cross(p[i], p[i + 1], p[i + 2]); 54 if (tmp > EPS) return; 55 else if (tmp < -EPS) 56 { 57 reverse(p, p + N); 58 return; 59 } 60 } 61 } 62 63 //计算C点到线段AB的最短距离 64 inline double point_to_line(Point A, Point B, Point C) 65 { 66 if (dist(A, B) < EPS) return dist(B, C); 67 if (multi(A, B, C) < -EPS) return dist(A, C); 68 if (multi(B, A, C) < -EPS) return dist(B, C); 69 return fabs(cross(A, B, C) / dist(A, B)); 70 } 71 72 //求一条线段的两端点到另外一条线段的距离，反过来一样，共4种情况 73 inline double line_to_line(Point A, Point B, Point C, Point D) 74 { 75 return min(min(point_to_line(A, B, C), point_to_line(A, B, D)), min(point_to_line(C, D, A), point_to_line(C, D, B))); 76 } 77 78 double solve(Point* P, Point* Q, int n, int m) 79 { 80 int yminP = 0, ymaxQ = 0; 81 for (int i = 0; i < n; ++i) if (P[i].y < P[yminP].y) yminP = i; // P上y坐标最小的顶点 82 for (int i = 0; i < m; ++i) if (Q[i].y > Q[ymaxQ].y) ymaxQ = i; // Q上y坐标最大的顶点 83 P[n] = P[0]; // 为了方便，避免求余 84 Q[m] = Q[0]; 85 double arg, ans = INF; 86 for (int i = 0; i < n; ++i) 87 { 88 while (arg = cross(P[yminP + 1], Q[ymaxQ + 1], P[yminP]) - cross(P[yminP + 1], Q[ymaxQ], P[yminP]) > EPS) ymaxQ = (ymaxQ + 1) % m; 89 ans = min(ans, line_to_line(P[yminP], P[yminP + 1], Q[ymaxQ], Q[ymaxQ + 1])); 90 yminP = (yminP + 1) % n; 91 } 92 return ans; 93 } 94 95 96 ///////////////////////////SubMain////////////////////////////////// 97 int main(int argc, char *argv[]) 98 { 99 #ifndef ONLINE_JUDGE 100 freopen("in.txt", "r", stdin); 101 freopen("out.txt", "w", stdout); 102 #endif 103 int N, M; 104 while (~scanf("%d%d", &N, &M) && N + M) 105 { 106 for (int i = 0; i < N; ++i) 107 { 108 scanf("%lf%lf", &P[i].x, &P[i].y); 109 } 110 for (int i = 0; i < M; ++i) 111 { 112 scanf("%lf%lf", &Q[i].x, &Q[i].y); 113 } 114 anticlockwise_sort(P, N); 115 anticlockwise_sort(Q, M); 116 printf("%.5lf\n", solve(P, Q, N, M)); 117 } 118 #ifndef ONLINE_JUDGE 119 fclose(stdin); 120 fclose(stdout); 121 system("out.txt"); 122 #endif 123 return 0; 124 }
