PAT1004 Counting Leaves（30）（BFS，DFS，樹的層序遍歷）

類型：樹的遍歷，dfs

題目：

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

Input

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.

Output

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

分析題目

可憐英語不好，題目都看不懂，先查一下一些單詞
family hierarchy家譜 pedigree tree譜系樹
For the sake of simplicity爲了簡單起見
seniority排行

好了單詞查完了，大概瞭解了題目意思，輸入端首先第一行給一個N表示樹的總節點個數，第二個數M表示non-leaf node個數，後面每一行就表示所有non-leaf的結構，最終輸出每一層葉子節點的個數

代碼

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
vector<int> v[100];
int book[100];
int maxdepth = -1;
void dfs(int index, int depth)
{
    if(v[index].size() == 0)
    {
        maxdepth = max(depth, maxdepth);
        book[depth]++;
        return;
    }
    for(int i=0; i<v[index].size(); i++)
    {
        dfs(v[index][i], depth+1);
    }
}

int main()
{
    int N, M;
    cin >> N >> M;
    for(int i = 0; i<M; i++)
    {
        int index, childnum;
        cin >> index >> childnum;
        for(int j = 0; j<childnum; j++)
        {
            int child;
            cin >> child;
            v[index].push_back(child);
        }
    }
    dfs(1, 0);
    int k = 0;
    for(k=0; k<maxdepth; k++)
    {
        cout << book[k] << " ";
    }
    cout << book[k];
    return 0;
}

總結

其實是一道簡單的dfs題目，複習了一下dfs的操作，參考了柳神的代碼

還可以~