openjudge_2.5基本算法之搜索_1388:Lake Counting

题目

1388:Lake Counting
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总时间限制: 1000ms 内存限制: 65536kB
描述
Due to recent rains, water has pooled in various places in Farmer John’s field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water (‘W’) or dry land (‘.’). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John’s field, determine how many ponds he has.
输入

  • Line 1: Two space-separated integers: N and M

  • Lines 2…N+1: M characters per line representing one row of Farmer John’s field. Each character is either ‘W’ or ‘.’. The characters do not have spaces between them.
    输出

  • Line 1: The number of ponds in Farmer John’s field.
    样例输入
    10 12
    W…WW.
    .WWW…W

`std::counting_semaphore` 是 C++20 新增的同步原语之一,用于控制多个线程间的访问。它是一个计数信号量,可以用来限制同时访问某个资源的线程数量。在类的成员中使用 `std::counting_semaphore` 与在其他地方使用它并没有本质的区别,只需要在类的定义中声明一个 `std::counting_semaphore` 类型的成员即可。 以下是一个简单的示例代码: ```c++ #include <semaphore> #include <thread> #include <iostream> class Example { public: Example() : sema_(2) {} void do_something() { sema_.acquire(); std::cout << "Doing something..." << std::endl; std::this_thread::sleep_for(std::chrono::seconds(1)); sema_.release(); } private: std::counting_semaphore<2> sema_; }; int main() { Example e; std::thread t1(&Example::do_something, &e); std::thread t2(&Example::do_something, &e); std::thread t3(&Example::do_something, &e); t1.join(); t2.join(); t3.join(); return 0; } ``` 在这个例子中,`Example` 类中定义了一个 `std::counting_semaphore<2>` 类型的成员 `sema_`,用于控制同时访问 `do_something` 函数的线程数量。在 `do_something` 函数中,线程首先需要调用 `acquire()` 函数获取信号量,如果当前已经有两个线程在访问,则该线程会被阻塞,直到有一个线程调用了 `release()` 函数释放了信号量。在主函数中,我们创建了三个线程来同时访问 `do_something` 函数,由于信号量的数量是 2,因此最多只有两个线程能够同时访问,第三个线程需要等待前面的线程释放信号量后才能继续执行。
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