标签： OJ

When register on a social network, you are always asked to specify your hobbies in order to find some potential friends with the same hobbies. A “social cluster” is a set of people who have some of their hobbies in common. You are supposed to find all the clusters.
Input Specification:
Each input file contains one test case. For each test case, the first line contains a positive integer N (<=1000), the total number of people in a social network. Hence the people are numbered from 1 to N. Then N lines follow, each gives the hobby list of a person in the format:
Ki: hi[1] hi[2] … hi[Ki]
where Ki (>0) is the number of hobbies, and hi[j] is the index of the j-th hobby, which is an integer in [1, 1000].
Output Specification:
For each case, print in one line the total number of clusters in the network. Then in the second line, print the numbers of people in the clusters in non-increasing order. The numbers must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
8
3: 2 7 10
1: 4
2: 5 3
1: 4
1: 3
1: 4
4: 6 8 1 5
1: 4
Sample Output:
3
4 3 1

题目大意：有n个人，每个人喜欢k个活动，如果两个人有任意一个活动相同，就称为他们处于同一个社交网络。求这n个人一共形成了多少个社交网络。
分析：并查集。先写好init、findFather、Union。
0. 每个社交圈的结点号是人的编号，而不是课程。课程是用来判断是否处在一个社交圈的。
1. course[t]表示任意一个喜欢t活动的人的编号。如果当前的课程t，之前并没有人喜欢过，那么就course[t] = i，i为它自己的编号，表示i为喜欢course[t]的一个人的编号
2. course[t]是喜欢t活动的人的编号，那么findFather(course[t])就是喜欢这个活动的人所处的社交圈子的根结点，合并根结点和当前人的编号的结点i。即Union(i, findFather(course[t]))，把它们处在同一个社交圈子里面
3. isRoot[i]表示编号i的人是不是它自己社交圈子的根结点，如果等于0表示不是根结点，如果不等于0，每次标记isRoot[findFather(i)]++，那么isRoot保存的就是如果当前是根结点，那么这个社交圈里面的总人数
4. isRoot中不为0的编号的个数cnt就是社交圈圈子的个数
5. 把isRoot从大到小排列，输出前cnt个，就是社交圈人数的从大到小的输出顺序

Programming Ability Test (PAT) is organized by the College of Computer Science and Technology of Zhejiang University. Each test is supposed to run simultaneously in several places, and the ranklists will be merged immediately after the test. Now it is your job to write a program to correctly merge all the ranklists and generate the final rank.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive number N (<=100), the number of test locations. Then N ranklists follow, each starts with a line containing a positive integer K (<=300), the number of testees, and then K lines containing the registration number (a 13-digit number) and the total score of each testee. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in one line the total number of testees. Then print the final ranklist in the following format:

registration_number final_rank location_number local_rank

The locations are numbered from 1 to N. The output must be sorted in nondecreasing order of the final ranks. The testees with the same score must have the same rank, and the output must be sorted in nondecreasing order of their registration numbers.

Sample Input:
2
5
1234567890001 95
1234567890005 100
1234567890003 95
1234567890002 77
1234567890004 85
4
1234567890013 65
1234567890011 25
1234567890014 100
1234567890012 85
Sample Output:
9
1234567890005 1 1 1
1234567890014 1 2 1
1234567890001 3 1 2
1234567890003 3 1 2
1234567890004 5 1 4
1234567890012 5 2 2
1234567890002 7 1 5
1234567890013 8 2 3
1234567890011 9 2 4

题目大意：有n个考场，每个考场有若干数量的学生，给出每个考场中考生的编号和分数，要求算排名，输出所有考生的编号、排名、考场号、考场内排名
分析：先按照考场内排名 然后赋值给总数组fin，然后总排名，最后输出。注意相同的分数情况下按照学号的从小到大排列，但是他们的排名应该是一样的数字～

An AVL tree is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. Figures 1-4 illustrate the rotation rules.
Now given a sequence of insertions, you are supposed to tell the root of the resulting AVL tree.

Input Specification:
Each input file contains one test case. For each case, the first line contains a positive integer N (<=20) which is the total number of keys to be inserted. Then N distinct integer keys are given in the next line. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print ythe root of the resulting AVL tree in one line.
Sample Input 1:
5
88 70 61 96 120
Sample Output 1:
70
Sample Input 2:
7
88 70 61 96 120 90 65
Sample Output 2:
88

题目大意：AVL树是自平衡二叉搜索树。 在AVL树中，任何节点的两个子子树的高度最多相差一个；如果在任何时候它们相差多于一个，则重新平衡以恢复此属性。 图1-4说明了旋转规则～现在给出一系列插入，要求输出根节点的值～

分析：写出建AVL（平衡二叉搜索树）的代码模版即可，rotateLeft表示左旋，rotateRight表示右旋，rotateLeftRight表示先左旋后右旋，rotateRightLeft表示先右旋后左旋，getHeight表示获取传入结点的子树的高度，insert表示插入建树的过程，如果root为空，直接新建结点插入即可～如果当前要插入的值小于root->val，则插入root的左子树；如果当前要插入的值大于root->val，则插入root的右子树～如果插入后左右子树高度差大于1，再根据值的大小比较进行旋转调整使树平衡～插入完成后返回root指针赋值给main函数里的root～最后输出root的val值～

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
Given the structure of a binary tree and a sequence of distinct integer keys, there is only one way to fill these keys into the tree so that the resulting tree satisfies the definition of a BST. You are supposed to output the level order traversal sequence of that tree. The sample is illustrated by Figure 1 and 2.


Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100) which is the total number of nodes in the tree. The next N lines each contains the left and the right children of a node in the format “left_index right_index”, provided that the nodes are numbered from 0 to N-1, and 0 is always the root. If one child is missing, then -1 will represent the NULL child pointer. Finally N distinct integer keys are given in the last line.

Output Specification:

For each test case, print in one line the level order traversal sequence of that tree. All the numbers must be separated by a space, with no extra space at the end of the line.

Sample Input:
9
1 6
2 3
-1 -1
-1 4
5 -1
-1 -1
7 -1
-1 8
-1 -1
73 45 11 58 82 25 67 38 42
Sample Output:
58 25 82 11 38 67 45 73 42

题目大意：给出一棵二叉搜索树（给出每个结点的左右孩子），且已知根结点为0，求并且给出应该插入这个二叉搜索树的数值，求这棵二叉树的层序遍历
分析：1. 用结构体data，left，right表示这棵树的结构，a数组存树的信息，b数组存这棵树节点的所有data，根据输入可知树a[i]的left和right～
2. 因为是二叉搜索树，所以中序遍历这棵树得到的结点顺序应该是给出的数值序列从小到大的排列顺序，所以把数值序列排序后，可以在中序遍历的时候直接赋值当前a[root].data～同时可得知树的最大层数maxLevel的值～
3. 二维数组v用来存储每一层的结点下标，一共有0到maxLevel层。用for循环从0开始一层层遍历v，就可以得到下一层的l和r，遍历过程中可以输出每个结点对应的data值a[v[i][j]].data～

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: “Push X” where X is the index of the node being pushed onto the stack; or “Pop” meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1
题目大意：用栈的形式给出一棵二叉树的建立的顺序，求这棵二叉树的后序遍历
分析：栈实现的是二叉树的中序遍历（左根右），而每次push入值的顺序是二叉树的前序遍历（根左右），所以该题可以用二叉树前序和中序转后序的方法做～
root为当前子树的根结点在前序pre中的下标，start和end为当前子树的最左边和最右边的结点在中序in中的下标。用i找到当前子树的根结点root在中序中的下标，然后左边和右边就分别为当前根结点root的左子树和右子树。递归实现～
Update：Github用户littlesevenmo给我发issue提出题目并没有说所有节点的值互不相同。因此，在有多个节点的值相同的情况下，之前的代码会输出错误的结果，所以修改后的代码中添加了key作为索引，前中后序中均保存索引值，然后用value存储具体的值，修改后的代码如下：