PAT A1004 Counting Leaves(30)[数叶子结点，树]

PAT | AcWing

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

题意

找出树中每一层叶子结点数

思路1DFS

• 遍历到叶子结点将对应深度处的数量加1，同时更新最大深度以便于最后的输出
• 此处采用邻接表存储图，v[i].size() == 0 表示无孩子即为叶节点，否则从上到下遍历所有的孩子节点，传入的深度 depth+1

代码1

#include <iostream>
#include <vector>
using namespace std;
const int N = 110;
vector<int> v[N];  // 存储图
int d[N];   // 深度叶子个数
int maxdepth;

void dfs(int u, int depth)
{
    if(!v[u].size())  // 叶子结点
    {
        d[depth]++;
        maxdepth = max(maxdepth, depth);
        return;
    }
    for(int i = 0; i < v[u].size(); i++)
        dfs(v[u][i], depth+1);
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    while(m--)
    {
        int id, k, ch;
        scanf("%d%d", &id, &k);
        while(k--)
        {
            scanf("%d", &ch);
            v[id].push_back(ch);
        }
    }
    dfs(1, 0);
    printf("%d", d[0]);
    for(int i = 1; i <= maxdepth; i++)
        printf(" %d", d[i]);
    puts("");
    return 0;
}

思路2

代码2