用C语言分治法解决凸包问题
时间: 2024-01-05 16:03:08 浏览: 31
凸包问题是计算凸多边形的算法问题,常见的解决方法是使用 Graham 扫描算法和快速凸包算法。其中,快速凸包算法使用了分治思想,可以在 O(nlogn) 的时间复杂度内解决问题。
下面是用 C 语言实现的快速凸包算法:
```c
#include <stdio.h>
#include <stdlib.h>
struct point {
int x;
int y;
};
// 按照 x 坐标排序
int cmp(const void* a, const void* b) {
struct point* p1 = (struct point*)a;
struct point* p2 = (struct point*)b;
return (p1->x - p2->x);
}
// 计算叉积
int cross(struct point a, struct point b, struct point c) {
return ((b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x));
}
void quick_hull(struct point* points, int n, struct point p1, struct point p2, int side, struct point* hull, int* count) {
int i, pos = -1;
int max_dist = 0;
int dist;
// 找到到直线 p1-p2 最远的点
for (i = 0; i < n; i++) {
dist = cross(p1, p2, points[i]);
if (cross(p1, p2, points[i]) == side && abs(dist) > max_dist) {
max_dist = abs(dist);
pos = i;
}
}
// 如果没有点在凸包外,则将 p1 和 p2 加入凸包
if (pos == -1) {
hull[*count] = p1;
(*count)++;
hull[*count] = p2;
(*count)++;
return;
}
// 递归处理左右两个部分
quick_hull(points, n, points[pos], p1, -cross(points[pos], p1, p2), hull, count);
quick_hull(points, n, points[pos], p2, -cross(points[pos], p2, p1), hull, count);
}
void print_hull(struct point* hull, int n) {
int i;
for (i = 0; i < n; i++) {
printf("(%d, %d)\n", hull[i].x, hull[i].y);
}
}
int main() {
int n, i, count = 0;
struct point points[100], hull[100], p1, p2;
printf("Enter the number of points: ");
scanf("%d", &n);
printf("Enter the coordinates of each point:\n");
for (i = 0; i < n; i++) {
scanf("%d %d", &points[i].x, &points[i].y);
}
// 按照 x 坐标排序
qsort(points, n, sizeof(struct point), cmp);
// 左边部分
p1 = points[0];
p2 = points[n - 1];
hull[count] = p1;
count++;
quick_hull(points, n, p1, p2, 1, hull, &count);
// 右边部分
p1 = points[n - 1];
p2 = points[0];
quick_hull(points, n, p1, p2, 1, hull, &count);
printf("The points in the convex hull are:\n");
print_hull(hull, count);
return 0;
}
```
该算法的时间复杂度是 O(nlogn),空间复杂度是 O(n)。