PAT 甲级 1004 Counting Leaves (30)(30 分)

本博客介绍了一个算法问题,即在一个家族树结构中,统计每级中没有子女的成员数量。输入包括节点数量、非叶节点数量及各非叶节点的子节点信息,输出则为从根节点开始,各级的无子女成员数。

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1004 Counting Leaves (30)(30 分)

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
# include <iostream>
# include <queue>
# include <vector>
# include <algorithm>
using namespace std;
const int maxn = 111;

int n, m, node, k, id;
int leaf_node[maxn], maxdepth = -1;
vector<int> v[maxn];

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

int main(){
//	freopen("C:\\1.txt", "r", stdin);
	fill(leaf_node, leaf_node + maxn, 0);
	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", &id);
			v[node].push_back(id);
		}
	}
	dfs(1, 0);
	printf("%d", leaf_node[0]);
	for(int i = 1; i <= maxdepth; i++)
		printf(" %d", leaf_node[i]);
	return 0l;
}

 

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