实现平面礼品包装算法的 C++ 程序
我们开发一个 C++ 程序,在两个维度上实现礼品包装算法。礼品包装算法是用于计算给定点集凸包的算法。
算法
Begin function convexHull() to find convex hull of a set of n points: There must be at least three points. Initialize the result. Find the leftmost point. Starting from leftmost point, keep moving counterclockwise until reach the start point again. Print the result. End
示例代码
#include <iostream> using namespace std; #define INF 10000 struct P { int x; int y; }; int orient(P a, P b, P c) { int v = (b.y - a.y) * (c.x - b.x) - (b.x - a.x) * (c.y - b.y); if (v == 0) return 0; // colinear return (v >0) ? 1 : 2; // clock or counterclock wise } void convexHull(P points[], int m) { if (m < 3)//at least three points required return; int n[m]; for (int i = 0; i < m; i++) n[i] = -1; int l = 0;//initialize result. for (int i = 1; i < m; i++) if (points[i].x < points[l].x) l = i; //find left most point int p = l, q; do { q = (p + 1) % m; for (int i = 0; i < m; i++) if (orient(points[p], points[i], points[q]) == 2) q = i; n[p] = q; p = q; } while (p != l); for (int i = 0; i < m; i++) { if (n[i] != -1) cout << "(" << points[i].x << ", " << points[i].y << ")\n"; } } int main() { P points[] = {{0, 4}, {2, 1}, {2, 3}, {4, 1}, {3, 0}, {1, 1}, {7, 6}}; cout << "The points in the convex hull are: "; int n = sizeof(points) / sizeof(points[0]); convexHull(points, n); return 0; }
输出
The points in the convex hull are: (0, 4) (4, 1) (3, 0) (1, 1) (7, 6)
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