g -> Pair.of

时间: 2024-03-21 21:43:24 浏览: 21
`Pair.of()`是一个静态方法,用于创建一个`Pair`对象,表示由两个元素组成的有序对。`Pair`是一个通用的数据结构,在Java中没有内置的`Pair`类,但可以通过第三方库(如Apache Commons Lang)来创建`Pair`对象。 `Pair.of()`的基本用法如下: ```java Pair<T, U> pair = Pair.of(T t, U u); ``` 其中,`T`和`U`分别表示`Pair`对象的两个元素的类型,`t`和`u`分别表示`Pair`对象的两个元素的值。 例如,我们有一个`Map`对象,将其中的键值对按照值进行排序,并将排序后的结果放入一个`List`中: ```java Map<String, Integer> map = new HashMap<>(); map.put("Alice", 18); map.put("Bob", 21); map.put("Charlie", 16); map.put("David", 20); List<Pair<String, Integer>> sortedList = map.entrySet().stream() .sorted(Map.Entry.comparingByValue()) .map(g -> Pair.of(g.getKey(), g.getValue())) .collect(Collectors.toList()); System.out.println(sortedList); ``` 在上述代码中,`map.entrySet().stream()`将`Map`转换成了一个`Stream`对象,`Map.Entry.comparingByValue()`用于按照值进行排序,`Pair.of(g.getKey(), g.getValue())`将每个键值对转换成一个`Pair`对象,最后通过`Collectors.toList()`将结果收集到一个`List`中。 运行上述代码,输出结果为: ``` [(Charlie, 16), (Alice, 18), (David, 20), (Bob, 21)] ``` 可以看到,最终得到了一个按照值进行排序的键值对列表。

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3.4 Pair Interaction Feature The interaction pattern between two individuals is encoded by a spatial descriptor with view invariant relative pose encoding. Given the 3D locations of two individual detec- tions zi,zj and two pose features pi,pj, we represent the pairwise relationship using view normalization, pose co-occurrence encoding, semantic compression and a spatial histogram (see Fig. 5 for illustration). The view normalization is performed by rotating the two people in 3D space by θ with respect to their midpoint, making their connecting line perpendicular to the cam- era view point. In this step, the pose features are also shifted accordingly (e.g. if θ = 45‘, shift 1 dimension with a cycle). Then, the co-occurrence feature is obtained by building a 2-dimensional matrix in which each element (r, c) corresponds to min(pi(r), pj (c)). Although the feature is view invariant, there are still elements in the matrix that deliver the same semantic concepts (e.g. left-left and right-right). To reduce such unnecessary variance and obtain a compact representation, we perform another transformation by multiplying a semantic compression matrix Sc to the vector form of the co-occurrence feature. The matrix Sc is learned offline by enumerating all possible configurations of view points and grouping the pairs that are equivalent when rotated by 180 degrees. Finally, we obtain the pair interaction descriptor by building a spatial histogram based on the 3D distance between the two (bin centers at 0.2, 0.6, 2.0 and 6.5 m). Here, we use linear interpolation similarly to contextual feature in Sec. 3.3. Given the interac- tion descriptor for each pair, we represent the interaction feature φxx(xi,xj) using the confidence value from an SVM classifier trained on a dictionary of interaction labels Y.什么意思

补全以下代码private String cid;// Course id, e.g., CS110. private String name;// Course name, e.g., Introduce to Java Programming. private Integer credit;// Credit of this course private GradingSchema gradingSchema; //Grading schema of this course // enum GradingSchema{FIVE_LEVEL, PASS_FAIL} private Integer capacity;// Course capacity. private Integer leftCapacity;// Course capacity left. You should update the left capacity when enrolling students. private Set<Timeslot> timeslots;// One course may have one or more timeslots. e.g., a lecture in Monday's 10:20-12:10, and a lab in Tuesday's 14:00-15:50. public Course(String cid, String name, Integer credit, GradingSchema gradingSchema, Integer capacity) // constructor public void addTimeslot(Timeslot timeslot) //Record a timeslot for this course private Integer id;// A unique student id, should be an 8-digit integer: Undergraduates' ids should start with 1; Postgraduates' ids should start with 3. e.g., 12213199. private String name;// Student’s name private Map<Course, Grade> courses;// Enrolled courses, using Map structure to store course and its grade as a pair. Grade is an enum type enum Grade{PASS,FAIL,A,B,C,D,F}with an attribute: Double gradePoint protected Student(Integer id, String name) // constructor public abstract boolean canGraduate() // Checks if this student satisfies all the graduating conditions. Hint: you are allowed to change this abstract method into non-abstract to check if the student satisfies the common graduation conditions. public void enroll(Course course) // Tries to enroll the course, do some checks before enrolling. public void recordGrade(Course course, Grade grade)// Records the grade of a course that is current learning. public double getGpa() // Calculates the GPA for this student. public UndergraduateStudent(Integer id, String name)// constructor public boolean canGraduate() //Additional graduating conditions for undergraduate students public PostgraduateStudent(Integer id, String name)// constructor public boolean canGraduate() //Additional graduating conditions for postgraduate students

Solve the problem with c++ code, and give your code: Ack Country has N cities connected by M one-way channels. The cities occupied by the rebels are numbered 1, while the capital of Ack country is numbered N. In order to reduce the loss of effective force, you are permitted to use self-propelled bombers for this task. Any bomber enters the capital, your job is done. This seems simple enough, but the only difficulty is that many cities in Ack Country are covered by shields. If a city is protected by a shield, all shield generators that maintain the shield need to be destroyed before the bomber can enter or pass through the city. Fortunately, we know the cities where all the shield generators are located, and which cities' shields are being charged. If the bomber enters a city, all of its shield generators can be destroyed instantly. You can release any number of Bombermen and execute any command at the same time, but it takes time for bombermen to pass through the roads between cities. Please figure out how soon you can blow up Ack Nation's capital. The clock is ticking. Input: Two positive integers N,M in the first row. The next M lines, each with three positive integers, indicate that there is a road leading from the city to the city. It takes w time for the bomber to cross this road. Then N lines, each describing a city's shield. The first is a positive integer n, representing the number of shield generators that maintain shields in the city. Then n_i city numbers between 1 and N, indicating the location of each shield generator. In other words, if your bomber needs to enter the city, the bomber needs to enter all the entered cities in advance. If n_i=0, the city has no shields. Guarantee n_i=0.Output: a positive integer, the minimum time to blow up the capital. e.g., Input: 6 6 1 2 1 1 4 3 2 3 3 2 5 2 4 6 2 5 3 2 0 0 0 1 3 0 2 3 5, Output: 6.

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