PAT 甲级 1004 Counting Leaves（30）

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

Experiential Summing-up

题意呢就是给出一棵树，求每一层有多少个叶子结点。从根节点开始遍历，遍历到孩子结点时将当前层数 depth 的 book[depth]++。

Accepted Code

#include<bits/stdc++.h>
using namespace std;

int n, m, node, c, t;
int maxdepth = -1;
int book[100];//树的第depth层共有book[depth]个叶子结点
vector<int> v[100];

void dfs(int index, int depth)
{
    if(v[index].size() == 0)
    {
        book[depth] ++;
        maxdepth = max(maxdepth, depth);
        return;
    }
    for(int i = 0; i < v[index].size(); i ++)
        dfs(v[index][i], depth + 1);
}

int main()
{
    cin >> n >> m;
    for(int i = 0; i < m ; i ++)
    {
        cin >> node >> t;
        for(int j = 0; j < t; j ++)
        {
            cin >> c;
            v[node].push_back(c);
        }
    }
    dfs(1, 0);
    cout << book[0];
    for(int i = 1; i <= maxdepth; i ++)
        cout << " " << book[i];
    return 0;
}