1004 Counting Leaves

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, the number of nodes in a tree, and M (<), 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：

给一棵树，寻找每一层的叶子节点数。输入的第一行的两个数是结点的总数和非叶子结点的个数。

 

Code：

#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>

using namespace std;

vector<int> v[100];
int book[100], maxdepth = -1;

void dfs(int index, int depth) {
    if (v[index].size() == 0) {             //是不是叶子节点
        book[depth]++;
        maxdepth = max(maxdepth, depth);
        return ;
    }
    for (int i = 0; i < v[index].size(); ++i)   // 遍历该节点的子节点
        dfs(v[index][i], depth + 1);        
}

int main() {

    int n, m, k, node, c;
    scanf("%d %d", &n, &m);
    for (int i = 0; i < m; ++i) {
        scanf("%d %d", &node, &k);
        for (int j = 0; j < k; ++j) {
            scanf("%d", &c);
            v[node].push_back(c);
        }
    }

    dfs(1, 0);

    printf("%d", book[0]);
    for(int i = 1; i <= maxdepth; ++i)
        printf(" %d", book[i]);

    return 0;
}

 

参考：

https://www.liuchuo.net/archives/2229