已知三维数据坐标(x,y,z)存储在文件中。求三维点云的最近的两个点的坐标和距离 输入:文件名。 输出:最近的两个点的坐标和距离。并用c语言表达
时间: 2023-05-30 21:06:27 浏览: 61
题目描述不够清晰,以下是一种可能的解法。
假设文件中每一行包含一个点的坐标(x,y,z),先读入所有点到一个数组中。然后对于每一个点,计算它与其他所有点的距离,找到距离最近的两个点,记录下它们的坐标和距离。最后输出这两个点的坐标和距离。
具体的实现可以参考下面的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX_POINTS 10000
struct Point {
double x, y, z;
};
double distance(const struct Point* p1, const struct Point* p2) {
double dx = p1->x - p2->x;
double dy = p1->y - p2->y;
double dz = p1->z - p2->z;
return sqrt(dx*dx + dy*dy + dz*dz);
}
int main(int argc, char* argv[]) {
if (argc != 2) {
printf("Usage: %s <filename>\n", argv[0]);
return 1;
}
FILE* fp = fopen(argv[1], "r");
if (!fp) {
printf("Failed to open file %s\n", argv[1]);
return 1;
}
struct Point points[MAX_POINTS];
int n = 0;
while (fscanf(fp, "%lf,%lf,%lf", &points[n].x, &points[n].y, &points[n].z) == 3) {
n++;
if (n >= MAX_POINTS) {
printf("Too many points, max allowed is %d\n", MAX_POINTS);
return 1;
}
}
fclose(fp);
double min_dist = INFINITY;
struct Point* min_p1 = NULL;
struct Point* min_p2 = NULL;
for (int i = 0; i < n; i++) {
for (int j = i+1; j < n; j++) {
double dist = distance(&points[i], &points[j]);
if (dist < min_dist) {
min_dist = dist;
min_p1 = &points[i];
min_p2 = &points[j];
}
}
}
printf("closest points: (%.2f,%.2f,%.2f) and (%.2f,%.2f,%.2f)\n",
min_p1->x, min_p1->y, min_p1->z, min_p2->x, min_p2->y, min_p2->z);
printf("distance: %.2f\n", min_dist);
return 0;
}
```
假设文件中有以下内容:
```
1.0,2.0,3.0
4.0,5.0,6.0
7.0,8.0,9.0
```
运行程序:
```
$ ./nearest_points points.txt
closest points: (1.00,2.00,3.00) and (4.00,5.00,6.00)
distance: 5.20
```
表示点(1,2,3)和点(4,5,6)之间的距离最近,为5.20。