PAT甲级 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 . 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

结尾无空行

Sample Output:

0 1

结尾无空行

#include <iostream>
#include <vector>
using namespace std;

int maxlevel = 1;
int leaf[100] = {
    
    0};
vector<int> P[100];

void DFS(int parent, int level)
{
    
    
    maxlevel = max(maxlevel, level);
    // 如果该点没有孩子结点
    if (!P[parent].size())
    {
    
    
        leaf[level]++;
        return;
    }
    // 依次搜索该点的每一个孩子结点
    for (int i = 0; i < P[parent].size(); i++)
    {
    
    
        DFS(P[parent][i], level + 1);
    }
}

int main()
{
    
    
    int N, M;
    cin >> N >> M;
    for (int i = 0; i < M; i++)
    {
    
    
        int parent, k, id;
        cin >> parent >> k;
        for (int j = 0; j < k; j++)
        {
    
    
            cin >> id;
            P[parent].push_back(id);
        }
    }
    DFS(1, 1);
    cout << leaf[1];
    for (int i = 2; i <= maxlevel; i++)
    {
    
    
        cout << " " << leaf[i];
    }
    return 0;
}

简单的深度优先搜索，需要注意的是数组存放孩子节点。每次递归要更新树的最大深度。