以下是一个可能的C++最优解代码示例:
#include <iostream>
#include <vector>
#include <algorithm>
struct Request {
int wrap;
double distance;
int index;
bool operator<(const Request& other) const {
return wrap < other.wrap || (wrap == other.wrap && distance < other.distance);
}
};
double calculateMoveTime(const Request& from, const Request& to) {
double time = 0.0;
if (from.wrap != to.wrap) {
// Change band
time += 4.0;
// Change wrap
time += 1.0;
}
// Move to anchor position
int anchorPosition = static_cast<int>(to.distance / 30.0) * 30;
if ((to.distance - anchorPosition) * (to.distance - from.distance) < 0 && from.wrap == to.wrap) {
// No need to change direction or move to anchor position
time += std::abs(to.distance - from.distance) / 2.0;
} else {
time += std::abs(anchorPosition - from.distance) / 8.0; // Move to the closest anchor
time += 2.5; // Penalty for changing direction
if (anchorPosition != to.distance)
time += std::abs(to.distance - anchorPosition) / 2.0; // Slowly move to the target position
}
return time;
}
std::vector<int> optimizeScheduler(std::vector<Request>& requests) {
std::sort(requests.begin(), requests.end());
std::vector<int> order(requests.size());
for (int i = 0; i < order.size(); ++i)
order[i] = i;
std::stable_sort(order.begin(), order.end(), [&](int a, int b) {
return calculateMoveTime(requests[a], requests[0]) < calculateMoveTime(requests[b], requests[0]);
});
return order;
}
int main() {
int n;
std::cin >> n;
std::vector<Request> requests(n);
for (int i = 0; i < n; ++i) {
std::cin >> requests[i].wrap >> requests[i].distance;
requests[i].index = i;
}
std::vector<int> optimalOrder = optimizeScheduler(requests);
for (int index : optimalOrder)
std::cout << index << " ";
std::cout << std::endl;
return 0;
}
这段代码中,我们首先定义了一个Request
结构体来表示请求的信息,并重载了小于运算符以便排序。然后,我们实现了calculateMoveTime
函数来计算从一个位置移动到另一个位置所需的时间。接下来,我们通过对请求进行排序并根据最优调度准则重新排列它们的顺序。最后,我们输出优化后的调度顺序。
请注意,这只是一个可能的解决方案示例,具体实现可能因问题要求、性能和其他因素而有所不同。
内容由零声教学AI助手提供,问题来源于学员提问