PAT-甲级-1004 Counting Leaves C++

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 0<N<100, the number of nodes in a tree, and M (<N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] ... ID[K]

where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID's of its children. For the sake of simplicity, let us fix the root ID to be 01.

The input ends with N being 0. That case must NOT be processed.

Output Specification:

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output 0 1 in a line.

Sample Input:

2 1
01 1 02

Sample Output:

0 1

题意：

题目意思是按照树的高度，从根节点（level 0）开始，从上往下输出没有叶子节点的个数。

分析：

用Tree来描述节点，用一个vector来存储所有节点。处理输入的过程中确定每一个节点的孩子节点的情况。然后用BFS遍历这棵树，确定每个节点的高度，最后输出。

代码：

//
// Created by yaodong on 2023/7/11.
//
#include <vector>
#include "iostream"
#include "cstring"
#include "queue"
class Tree{
public:
    int index;                // 在数组中的位置
    int num;                  // 孩子节点的个数
    int depth;                // 节点所处高度
    std::vector<int> child;    // 记录孩子节点的在数组中的位置—index
    Tree(int index){
        this->index = index;
        this->num = 0;
        this->depth = 0;
    }
    Tree(int index, int num, int childArray[]){
        this->index = index;
        this->num = num;
        for (int i = 0; i < num; ++i)
            this->child.push_back(childArray[i]);
    }
};

int main() {
    int n, m;
    std::cin >> n >> m;
    std::vector<Tree> treeVector;
    for (int i = 0; i < n; ++i) {
        treeVector.emplace_back(i);
    }
//    printf("%d\n", treeVector.size());
    for (int i = 0; i < m; ++i) {
        int parentId;
        std::cin >> parentId;
        Tree &curNode = treeVector[parentId-1];
        int k;
        std::cin >> k;
        curNode.num = k;
        for (int j = 0; j < k; ++j) {
            int temp;
            std::cin >> temp;
            curNode.child.push_back(temp-1);
        }
    }
    int noLeafDepth[n];
    memset(noLeafDepth, 0, sizeof(noLeafDepth));
    std::queue<int> q;        //用队列的性质来处理depth，从根节点0开始
    q.push(0);
    while(!q.empty()){
        int first = q.front();
        q.pop();
        Tree &t = treeVector[first];
        for (int i = 0; i < t.num; ++i) {
            int childIndex = t.child[i];
            treeVector[childIndex].depth = t.depth + 1;
            q.push(childIndex);
        }
    }
    int maxDepth = -1;
    for (int i = 0; i < n; ++i) {
        Tree t = treeVector[i];
        if(t.depth > maxDepth)
            maxDepth = t.depth;
        if(t.num == 0)
            noLeafDepth[t.depth]++;
    }
//    printf("maxDepth: %d\n", maxDepth);
    for (int i = 0; i <= maxDepth; ++i) {
        if(i == 0)
            printf("%d", noLeafDepth[i]);
        else
            printf(" %d", noLeafDepth[i]);
    }
}