用分治算法解决包含所有点的最小凸包问题的c语言代码

以下是使用分治算法解决包含所有点的最小凸包问题的C语言代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define MAX_POINTS 1000

typedef struct {
    double x;
    double y;
} point;

point points[MAX_POINTS];
int n;

int cmp(const void *a, const void *b) {
    point *p1 = (point *)a;
    point *p2 = (point *)b;

    if (p1->x < p2->x) {
        return -1;
    } else if (p1->x == p2->x) {
        if (p1->y < p2->y) {
            return -1;
        } else if (p1->y == p2->y) {
            return 0;
        } else {
            return 1;
        }
    } else {
        return 1;
    }
}

double cross(point p1, point p2, point p3) {
    return (p2.x - p1.x) * (p3.y - p1.y) - (p2.y - p1.y) * (p3.x - p1.x);
}

void merge(int l, int m, int r, point *temp) {
    int i, j, k;
    i = l;
    j = m + 1;
    k = 0;

    while (i <= m && j <= r) {
        if (points[i].x < points[j].x) {
            temp[k++] = points[i++];
        } else {
            temp[k++] = points[j++];
        }
    }

    while (i <= m) {
        temp[k++] = points[i++];
    }

    while (j <= r) {
        temp[k++] = points[j++];
    }

    for (i = l, j = 0; i <= r; i++, j++) {
        points[i] = temp[j];
    }
}

void find_convex_hull(int l, int r, point *hull) {
    if (l == r) {
        hull[0] = points[l];
        return;
    }

    int m = (l + r) / 2;
    point left_hull[MAX_POINTS], right_hull[MAX_POINTS];
    find_convex_hull(l, m, left_hull);
    find_convex_hull(m + 1, r, right_hull);

    int i, j, k;
    i = 0;
    j = 0;
    k = 0;

    while (i < m - l + 1 && j < r - m) {
        if (left_hull[i].x < right_hull[j].x) {
            hull[k++] = left_hull[i++];
        } else {
            hull[k++] = right_hull[j++];
        }
    }

    while (i < m - l + 1) {
        hull[k++] = left_hull[i++];
    }

    while (j < r - m) {
        hull[k++] = right_hull[j++];
    }

    int nh = k;

    for (i = 0; i < nh; i++) {
        points[i] = hull[i];
    }

    k = 1;

    for (i = 2; i < nh; i++) {
        while (k > 0 && cross(points[k - 1], points[k], points[i]) < 0) {
            k--;
        }
        points[++k] = hull[i];
    }

    int t = k;

    for (i = nh - 2; i >= 0; i--) {
        while (k > t && cross(points[k - 1], points[k], hull[i]) < 0) {
            k--;
        }
        points[++k] = hull[i];
    }

    if (r - l + 1 > n) {
        n = r - l + 1;
    }
}

int main() {
    int i;
    point hull[MAX_POINTS];
    point temp[MAX_POINTS];

    printf("Enter the number of points: ");
    scanf("%d", &n);

    printf("Enter the points:\n");

    for (i = 0; i < n; i++) {
        scanf("%lf %lf", &points[i].x, &points[i].y);
    }

    qsort(points, n, sizeof(point), cmp);
    find_convex_hull(0, n - 1, hull);

    printf("The minimum convex hull contains %d points:\n", n);

    for (i = 0; i < n; i++) {
        printf("(%lf, %lf)\n", points[i].x, points[i].y);
    }

    return 0;
}

该代码使用了分治算法来解决包含所有点的最小凸包问题。它首先对所有点按照 x 坐标进行排序，然后使用分治法将点集分为左右两部分，分别对左右两部分求解最小凸包。最后将左右两个凸包合并成一个凸包。

在求解最小凸包的过程中，该算法使用了一个双向扫描的方法来计算凸包的顶点。具体来说，从左到右扫描点集，将凸包的顶点依次加入到凸包中。然后从右到左扫描点集，将凸包的顶点依次加入到凸包中。最后得到的点集就是包含所有点的最小凸包。