C语言用分治法求二维坐标点集的最小近对值及下标的代码

在C语言中，使用分治算法解决二维坐标点集中找到两个最近点的问题通常涉及到计算每个点到所有其他点的距离，并维护一个局部的最接近距离和对应的下标。这里提供一个简化版的思路和伪代码示例，实际的完整代码会包含数据结构、排序和递归步骤：

// 假设points数组存储了二维坐标点集合，每个元素是一个struct Point，包含x和y坐标
typedef struct {
    int x;
    int y;
} Point;

// 计算两点之间的曼哈顿距离
int manhattan_distance(Point p1, Point p2) {
    return abs(p1.x - p2.x) + abs(p1.y - p2.y);
}

// 分治函数，用于查找全局最小距离和下标
void find_min_pair(int left, int right, Point points[], int n, int* min_dist, int* min_index1, int* min_index2) {
    if (left >= right) { // 如果只有一个元素，直接返回
        return;
    }
    
    int mid = (left + right) / 2;
    int dist_left = manhattan_distance(points[mid], points[left]);
    int closest_left = left;
    for (int i = left + 1; i < mid; i++) {
        if (manhattan_distance(points[i], points[left]) < dist_left) {
            dist_left = manhattan_distance(points[i], points[left]);
            closest_left = i;
        }
    }

    int dist_right = manhattan_distance(points[mid], points[right]);
    int closest_right = right;
    for (int i = right - 1; i > mid; i--) {
        if (manhattan_distance(points[i], points[right]) < dist_right) {
            dist_right = manhattan_distance(points[i], points[right]);
            closest_right = i;
        }
    }

    *min_dist = min(*min_dist, dist_left, dist_right); // 更新当前最小距离
    if (*min_dist == dist_left) {
        *min_index1 = closest_left;
        *min_index2 = mid;
    } else {
        *min_index1 = mid;
        *min_index2 = closest_right;
    }

    // 递归处理左右两部分
    find_min_pair(left, mid - 1, points, n, min_dist, min_index1, min_index2);
    find_min_pair(mid + 1, right, points, n, min_dist, min_index1, min_index2);
}

// 调用主函数并初始化结果变量
int main() {
    Point points[] = {/* 点集... */};
    int n = sizeof(points) / sizeof(points[0]);
    int min_dist = INT_MAX, min_index1 = 0, min_index2 = 0;
    find_min_pair(0, n - 1, points, n, &min_dist, &min_index1, &min_index2);
    printf("Minimum distance: %d\n", min_dist);
    printf("Indexes of the points: (%d, %d)\n", min_index1, min_index2);
    return 0;
}

注意：这个例子仅适用于求解单对最近点，如果需要求所有点对间的最小距离，你需要调整算法以适应这种情况。