强行做出 1004 Counting Leaves (30 分)

1004 Counting Leaves (30 分)

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

题目要求是按层输出每层的叶节点数量。

我的代码：

#include <iostream>
#include <vector>
using namespace std;
struct trNode {
    int pId;  //父节点ID
    vector<int> ch;
};
int main() {
    int n, m;
    cin >> n >> m;
    vector<trNode> tree(n + 1);  //节点起始索引是1
    for (int i = 0; i < m; i++) {
        int root, num;
        scanf("%d %d", &root, &num);
        for (int j = 0; j < num; j++) {
            int child;
            scanf("%d", &child);
            tree[child].pId = root;
            tree[root].ch.push_back(child);
        }
    }
    vector<int> layer{1};       //当前层所有节点ID
    while (layer.size() > 0) {  //若当前层节点数量不为0
        vector<int> temp;
        int non_leaf = 0;
        for (int i = 0; i < layer.size(); i++) {
            if (tree[layer[i]].ch.size() == 0)
                non_leaf++;
            else {  //下层节点由该层所有子节点组成
                temp.insert(temp.end(), tree[layer[i]].ch.begin(),
                            tree[layer[i]].ch.end());
            }
        }
        if (layer[0] != 1)
            printf(" ");
        printf("%d", non_leaf);
        layer = temp;
    }
    return 0;
}

写完后看看大家是怎么写的，结果都是用的DFS。感觉写法有点凑巧，所以记录一下。