PAT-A1004 Counting Leaves 【DFS】

A family hierarchy（层次） 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（同等的）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

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题目大意

• 一个测试用例
• 第一行两个参数，第一个参数n是节点的数量，第二个参数m是非叶子节点的数量
• 然后紧跟着m行，每行第一个参数是非叶子节点的id，第二个参数是该非叶子节点的孩子数量，紧接着就是孩子节点的id
• 要求输出每层非叶子节点的数量

思路

很明显，涉及到家庭成员份之情况是一个树操作。根据要求输出每层的非叶子节点，只要用到深搜（DFS）。

程序设计：首先使用Vector保存树的结构，然后使用dfs算法对其进行深搜。使用dfs需要注意的就是边界值，即什么时候应该退出递归，做一些小修改就完成了

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#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
using namespace std;
const int MAX_N = 100;
vector<int> children[MAX_N];    //使用vector保存树的结构
int maxFloor = 1;               //maxfloor这个变量很重要，此变量记录了整棵树的最大深度。
                                //按照逻辑，树的深度是多少，最后就要输出多少个数字。
int leafEveryFloor[MAX_N];      //储存深度为i的非叶子节点的数量，这里从1开始。
void dfs(int p, int floor){
    // 更新树的最大层次
    maxFloor = max(floor, maxFloor);     //更新maxfloor，取最大。
    
    const long l = children[p].size();   //某个节点的储存孩子的数组大小为0，即它本身就是叶节 
                                         //点
    if(l == 0){
        leafEveryFloor[floor]++; 
        return;
    }
    
    //这里floor++ 是为了下面的循环中不需要floor+1， 减少重复计算
    floor++; 
    for(int i = 0; i < l; i++){
        dfs(children[p][i], floor);
    }
}
int main() {
    // 读取输入
    int n, m, id, k, child;
    scanf("%d %d", &n, &m);
    while (m-- > 0) {
        scanf("%d %d", &id, &k);
        while (k-- > 0) {
            scanf("%d", &child);
            children[id].push_back(child);
        }
    }
    
    // 1号为根节点，默认第一根节点的深度是1
    dfs(1, 1);
    
    // 输出结果
    printf("%d", leafEveryFloor[1]);
    for(int i = 2; i <= maxFloor; i++){
        printf(" %d", leafEveryFloor[i]);
    }
    return 0;
}