帮我用C语言或者C++来写一个能解决上述类似问题的代码
时间: 2024-03-26 13:35:32 浏览: 11
以下是用C++实现的代码,可以解决上述问题:
```c++
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct Box {
int length, width, height;
};
struct Space {
int length, width, height;
int x, y, z;
};
bool cmp(const Box& a, const Box& b) {
return a.length > b.length;
}
vector<Box> boxes;
vector<Space> spaces;
void init() {
boxes.push_back({100, 100, 50});
boxes.push_back({100, 100, 50});
boxes.push_back({100, 100, 50});
boxes.push_back({150, 50, 20});
boxes.push_back({150, 50, 20});
boxes.push_back({100, 0, 0});
boxes.push_back({100, 0, 0});
boxes.push_back({0, 0, 100});
boxes.push_back({0, 0, 100});
boxes.push_back({0, 0, 100});
spaces.push_back({2000, 1000, 1000, 0, 0, 0});
}
void print_space(const Space& space) {
cout << "x: " << space.x << ", ";
cout << "y: " << space.y << ", ";
cout << "z: " << space.z << ", ";
cout << "length: " << space.length << ", ";
cout << "width: " << space.width << ", ";
cout << "height: " << space.height << endl;
}
void put_box(const Box& box, const Space& space) {
int x = space.x;
int y = space.y;
int z = space.z;
int length = box.length;
int width = box.width;
int height = box.height;
spaces.pop_back();
if (length <= space.length && width <= space.width && height <= space.height) {
Space next_space1 = {space.length - length, width, height, x + length, y, z};
Space next_space2 = {length, space.width - width, height, x, y + width, z};
Space next_space3 = {length, width, space.height - height, x, y, z + height};
if (next_space1.length > 0 && next_space1.width > 0 && next_space1.height > 0) {
spaces.push_back(next_space1);
}
if (next_space2.length > 0 && next_space2.width > 0 && next_space2.height > 0) {
spaces.push_back(next_space2);
}
if (next_space3.length > 0 && next_space3.width > 0 && next_space3.height > 0) {
spaces.push_back(next_space3);
}
cout << "Box (" << length << ", " << width << ", " << height << ") is put at ";
cout << "x: " << x << ", ";
cout << "y: " << y << ", ";
cout << "z: " << z << endl;
}
}
void solve() {
sort(boxes.begin(), boxes.end(), cmp);
while (!boxes.empty() && !spaces.empty()) {
Box box = boxes.back();
boxes.pop_back();
for (int i = 0; i < spaces.size(); i++) {
put_box(box, spaces[i]);
if (boxes.empty()) {
break;
}
}
}
if (!boxes.empty()) {
cout << "No enough space" << endl;
}
}
int main() {
init();
solve();
return 0;
}
```
代码的思路是:先将所有盒子按照长度从大到小排序,然后遍历所有盒子,在所有可用的空间中尝试放置盒子。如果当前盒子无法放置,则跳过该盒子,继续尝试下一个盒子。如果所有盒子都放置完毕,输出所有盒子的位置,否则输出 No enough space。
以上代码仅供参考,实际情况下可以根据具体需求进行修改。